Goyal Brothers Solutions for ICSE Class 9 Physics Chapter 10 Electricity And Magnetism 1

Unit I

Exercise 1

Question 1. There is a positively charged sphere A and negatively charged sphere B, such that they are brought in electrical contact by a copper wire. Answer the following questions :
(a) Which sphere is at higher potential before electrical contact on the basis of convention?
(b) Which sphere is at lower potential before electrical contact on the basis of convention?
(c) In which direction conventional current flows?
(d) In which direction electronic current flows?
(e) What is potential of the spheres after electrical contact?

Answer:
(a) On the basis of convention, positively charged sphere A is at higher potential before electrical contact.
(b) On the basis of convention, negatively charged sphere B is lower potential before electrical contact.
(c) Conventional current flows from sphere A to sphere B i.e. from a body at higher potential to the body at lower potential.
(d) Electronic current flows from sphere B to sphere A i.e. from a body at lower potential to the body at higher potential.
(e) After electrical contact, both the spheres will be at same potential.

In simple words: Positive charge is like a high hill and negative charge is like a valley. Conventional current flows "downhill" from positive to negative, while electrons actually travel "uphill" from negative to positive. Once connected, they level out to the same height.

📝 Teacher's Note: Use the analogy of water flowing from a higher tank to a lower tank to explain potential difference. The flow stops when the water levels (potentials) become equal.

🎯 Exam Tip: Remember the contradiction: "Conventional current" is a historical guess that is opposite to the actual path of "Electronic current" (electrons).

Question 2. (a) What do you understand by the term electric potential?
(b) Define electric potential in terms of energy spent.
(c) State the unit of electric potential and define it.

Answer:
(a) Electric potential : The amount of work done in moving a unit positive charge from infinity to a given point in an electric field is called electric potential.
(b) Electric potential : The amount of energy spent in moving a unit positive charge from infinity to a given point in an electric field is called electric potential.
(c) Volt is the SI unit of electric potential.
One volt : When one coulomb of charge is brought from infinity to a given point in an electric field, such that work done is one joule, then electric potential is said to be one volt.
OR
Electric potential is said to be one volt if one Joule of work is done in moving one coulomb of charge from infinity to a given point in an electric field.

In simple words: Electric potential is like "electrical pressure." One Volt is the amount of effort needed to push a standard amount of electricity from far away to a specific spot.

📝 Teacher's Note: Explain "Infinity" as a place so far away that the electric field of the target point has zero effect. This helps students grasp the absolute nature of potential.

🎯 Exam Tip: The definition of 1 Volt must include "1 Joule of work" and "1 Coulomb of charge" to be technically correct.

Question 3. (a) What do you understand by the term quantity of electric charge?
(b) State SI unit of electric charge and define it.
(c) How many electrons constitute one unit electric charge in SI system?

Answer:
(a) Quantity of electric charge : The number of charge (electrons) which drift from lower to higher potential is called quantity of charge.
(b) SI unit of electric charge is coulomb (C).
One coulomb : The quantity of electric charge which will deposit 0.00118 g of silver on the cathode, when passed through silver nitrate is called one coulomb.
(c) \( 6.25 \times 10^{18} \) electrons constitute one unit (1 C) electric charge in SI system.

In simple words: Charge is just a collection of many tiny electrons. One Coulomb is like a huge bucket that holds over 6 quintillion electrons.

📝 Teacher's Note: Use the "Silver Voltameter" definition for 1 Coulomb if students are studying electrolysis; otherwise, use the current-based definition (\( 1\text{ C} = 1\text{ A} \times 1\text{ s} \)).

🎯 Exam Tip: Remember the value \( 6.25 \times 10^{18} \). It is a standard numerical constant frequently used in electricity problems.

Question 4. (a) What do you understand by the term electric current?
(b) State and define the SI unit of electric current.
(c) State the relation between electric current; number of electrons moving in a circuit and time in seconds.

Answer:
(a) Electric current : The rate of flow of electric charge in an electric circuit is called electric current.
(b) Ampere (A) is the SI unit of electric current.
One ampere : When one coulomb charge flows through an electric circuit in one second, then the electric current flowing the circuit is said to be ampere.
(c) If Q is the charge (in coulombs) flowing through conductor in time t (in seconds) such that current I flows through the conductor then
Rate of flow of charge \( = Q/t \)
We know rate of flow of charge \( = I = \text{Electric current} \).

\( \implies I = Q/t \)

In simple words: Current is how fast electricity flows. If a road has many cars passing every second, it has a "high current" of traffic. Ampere measures this flow rate.

📝 Teacher's Note: Emphasize the word "rate." In physics, "rate" almost always means you are dividing by time (\( t \)).

🎯 Exam Tip: Units are just as important as the numbers. Always write 'A' for Amperes and 'C' for Coulombs in your final answers.

Question 5. How electric current flows in (i) solids, (ii) liquids?

Answer:
(i) Flow of electric current in solids : In solids, the positive charges are associated with atomic nuclei. As the nuclei are firmly packed and closely held by inter-atomic forces, therefore, positive charges cannot drift. On the other hand, negative charges (electrons) are not held firmly. Thus, when a potential difference, however small, is applied they start drifting from lower to higher potential. The continuous drift of electrons, through the body of a solid conductor constitutes the current.
(ii) Flow of electric current in liquids : Within a liquid no electrons move. However, when a negatively charged and a positively charged electrodes are placed in a liquid, it sets up an electric field. Under the influence of the electric field, the positively charged ions migrate towards the negatively charged electrode and vice versa. At the cathode the positively charged ions gain electrons. At the anode the negatively charged ions lose same number of electrons. Thus, in a way number of electrons given by the cathode is equal to the number of electrons accepted by the anode. To sum up, we can say that simultaneous movement and discharge of positive and negative ions in the opposite directions constitutes the current in the liquids.

In simple words: In wires, only tiny electrons can run through the solid metal blocks. In liquids, whole atoms with a charge (called ions) swim toward the battery terminals to carry the electricity.

📝 Teacher's Note: Clarify that in solids, "free electrons" are the charge carriers, while in liquids (electrolytes), "ions" (both positive and negative) are the carriers.

🎯 Exam Tip: A common MCQ asks for the charge carriers in different states. Solids = Electrons; Liquids = Ions; Gases = Ions and Electrons.

Question 6. (a) Define the term potential difference.
(b) How is potential difference related to work done and quantity of charge?

Answer:
(a) Potential difference : The amount of work done in moving a unit positive charge from one point to another point in an electric field is called potential difference.
(b) If Q \( = \) Charge moving from one point to another in an electric field.
W \( = \) Work done to move the charge Q from one point to another.
V \( = \) Potential difference between two points.
Then work done in moving Q units of charge \( = W \)
Work done in moving one unit of charge \( = W/Q \)
But work done in moving one unit of charge \( = \text{Potential difference} = V \)

\( \implies V = W/Q \)

In simple words: Potential difference is the "push" between two spots. If you do a lot of work to move a charge between two points, those points have a high potential difference (Voltage).

📝 Teacher's Note: Unlike absolute potential, potential difference is a relative measure. It's like measuring the height of a step rather than the height of the whole building.

🎯 Exam Tip: The formula \( V = W/Q \) is fundamental. Use it to check the units: \( 1\text{ Volt} = 1\text{ Joule} / 1\text{ Coulomb} \).

Practice Problems 1

Question 1. A charge of 5000 C flows through an electric circuit in 2 hours and 30 minutes. Calculate the magnitude of current in circuit.

Answer:
Charge \( = Q = 5000\text{ C} \)
Time \( = t = 2\text{ hour } 30\text{ minutes} \)

\( \implies t = 2 \times 60 + 30 = 150\text{ min} \)

\( \implies t = 150 \times 60 \text{ second} \)

\( \implies t = 9000\text{ second} \)
Electric current \( = I = ? \)

\( I = \frac{Q}{t} \)

\( I = \frac{5000}{9000} = 0.555\text{ A} \)

In simple words: We first turn the hours into seconds (the standard time unit). Then we divide the total amount of charge by that time to see how much flowed every second.

📝 Teacher's Note: Always convert time to seconds. This is where most students make their first mistake in electricity calculations.

🎯 Exam Tip: If your division results in a recurring decimal like 0.555..., round it to two or three decimal places as required by your marking scheme (e.g., 0.56 A).

Question 2. A charge of 8860 C flows through an electric circuit in 2 min and 40 s. Calculate the magnitude of current in circuit.

Answer:
Charge \( = Q = 8860\text{ C} \)
Time \( = t = 2\text{ min } 40\text{ s} \)

\( \implies t = (2 \times 60 + 40)\text{ s} \)

\( \implies t = 160\text{ s} \)
Electric current \( = I = ? \)

\( I = \frac{Q}{t} = \frac{8860}{160} \)

\( I = 55.375\text{ A} \)

In simple words: Just like the last problem, we find the total seconds and divide the charge by that time. This circuit has a very strong current!

📝 Teacher's Note: Reinforce the \( I = Q/t \) formula. Practice mental math for converting minutes to seconds (\( 120 + 40 \)).

🎯 Exam Tip: Don't leave your answer as a fraction. In physics, decimal answers with units are preferred.

Practice Problems 2

Question 1. A battery can supply a charge of \( 25 \times 10^4\text{ C} \). If the current is drawn from battery at the rate of 2.5 A, calculate the time in which battery will discharge completely.

Answer:
Charge \( = Q = 25 \times 10^4\text{ C} \)
Electric current \( = I = 2.5\text{ A} \)
Time \( = t = ? \)

\( I = \frac{Q}{t} \)

\( \implies t = \frac{Q}{I} \)

\( \implies t = \frac{25 \times 10^4}{2.5} = 10^5\text{ s} \)

\( t = 100000\text{ s} \)

In simple words: The battery has a huge "storage" of charge. Since we are taking it out at a steady rate, we divide the storage by the flow rate to find out how many seconds it lasts.

📝 Teacher's Note: Scientific notation can be scary for students. Show that \( 25 / 2.5 \) is simply 10, then add that 10 to the power of \( 10^4 \) to get \( 10^5 \).

🎯 Exam Tip: Use the triangle method to rearrange formulas: Q on top, I and t on the bottom. It helps you quickly find \( t = Q/I \).

Question 2. A dry cell can supply a charge of 800 C. If continuous current of 8.0 mA is drawn, calculate the time in which cell will discharge completely.

Answer:
Charge \( = Q = 800\text{ C} \)
Electric current \( = I = 8.0\text{ mA} \)

\( \implies I = 8 \times 10^{-3}\text{ A} \)
Time \( = t = ? \)

\( I = \frac{Q}{t} \)

\( \implies t = \frac{Q}{I} \)

\( \implies t = \frac{800}{8 \times 10^{-3}} = 100 \times 10^3 \)

\( t = 100000\text{ s} \)

In simple words: "mA" stands for milliamperes, which are tiny. Because the current is so small, even a small charge of 800 C takes a very long time to finish.

📝 Teacher's Note: Milli (m) means \( 10^{-3} \) or divide by 1000. Ensure students understand that smaller currents make batteries last longer.

🎯 Exam Tip: Be careful with prefix conversions! 1 mA \( = \) 0.001 A. Forgetting this will make your answer wrong by 1000 times.

Practice Problems 3

Question 1. Calculate the total number of electrons flowing through a circuit in 20 mins and 40 s, if a current of 40 \( \mu \text{A} \) flows through the circuit. [\( 1\text{ e}^- = 1.6 \times 10^{-19}\text{C} \)]

Answer:
Number of electrons \( = n = ? \)
Time \( = t = 20\text{ min } 40\text{ s} \)

\( \implies t = (20 \times 60 + 40)\text{ s} \)

\( \implies t = 1240\text{ s} \)
Electric current \( = I = 40 \mu \text{A} \)

\( \implies I = 40 \times 10^{-6}\text{ A} \)
Charge on one electron \( = e = 1.6 \times 10^{-19}\text{ C} \)
Charge \( = Q = ? \)

\( I = \frac{Q}{t} \)

\( \implies Q = It \)

\( \implies Q = 40 \times 10^{-6} \times 1240 \)

\( \implies Q = 49600 \times 10^{-6}\text{ C} \)

\( \implies Q = 4.96 \times 10^{-2}\text{ C} \)
Now \( Q = n \times (\text{charge on 1 electron}) = ne \)

\( \implies 4.96 \times 10^{-2} = n \times 1.6 \times 10^{-19} \)

\( \implies n = \frac{4.96 \times 10^{-2}}{1.6 \times 10^{-19}} \)

\( \implies n = 3.1 \times 10^{17}\text{ electrons} \)

In simple words: First we find the total charge using current and time. Then, knowing how tiny one electron is, we figure out how many of them it takes to make that total charge.

📝 Teacher's Note: This problem combines two formulas: \( I = Q/t \) and \( Q = ne \). This is a common way to test "Quantization of Charge."

🎯 Exam Tip: Prefix reminder: \( \mu \) (micro) means \( 10^{-6} \). Always check your scientific notation math at the end.

Question 2. \( 4 \times 10^{20} \) electrons flow through a circuit in 10 hours. Calculate magnitude of current. [\( 1\text{ e}^- = 1.6 \times 10^{-19}\text{ C} \)]

Answer:
Number of electrons \( = n = 4 \times 10^{20} \)
Time \( = t = 10\text{ hours} \)

\( \implies t = 10 \times 60 \times 60\text{ s} \)

\( \implies t = 36000\text{ s} \)
Charge on one electron \( = e = 1.6 \times 10^{-19}\text{ C} \)
Charge \( = Q = ne \)

\( \implies Q = 4 \times 10^{20} \times 1.6 \times 10^{-19} \)

\( \implies Q = 6.4 \times 10^1 \)

\( \implies Q = 64\text{ C} \)
Electric current \( = I = \frac{Q}{t} \)

\( I = \frac{64}{36000} = \frac{64}{36} \times 10^{-3}\text{ A} \)

\( I = 1.77 \times 10^{-3}\text{ A} \)

\( I = 1.77\text{ mA} \)

In simple words: We find out the total weight (charge) of all those electrons first. Then we divide that by the seconds in 10 hours to see the flow rate.

📝 Teacher's Note: Demonstrate how \( 10^{20} \times 10^{-19} \) results in just \( 10^1 \) (or 10). Powers of ten arithmetic is the key to these problems.

🎯 Exam Tip: If your answer is very small, like 0.00177 A, it's more professional to write it in milliamperes (1.77 mA).

Practice Problems 4

Question 1. What is the electrical potential at a point in an electric field when 24 J of work is done in moving a charge of 96 C from infinity?

Answer:
Electric potential \( = V = ? \)
Work, done \( = W = 24\text{ J} \)
Charge \( = Q = 96\text{ C} \).

\( V = \frac{W}{Q} \)

\( \implies V = \frac{24}{96} = \frac{1}{4} \)

\( V = 0.25\text{ V} \)

In simple words: Potential is work per charge. Since the amount of charge is much bigger than the work done, each unit of charge only gets a small bit of "push" (0.25 Volts).

📝 Teacher's Note: Use this to reinforce the definition of potential from Question 2. It is literally "Work divided by Charge."

🎯 Exam Tip: Always state the unit 'V' for Volts. In physics, a number without a unit is often marked as wrong.

Question 2. A charge of 75 C is brought from infinity to a given point in an electric field, when amount of work done is 3.75 J. Calculate the electrical potential at that point.

Answer:
Charge \( = Q = 75\text{ C} \)
Work done \( = W = 3.75\text{ J} \)
Electric potential \( = V = ? \)

\( V = \frac{W}{Q} \)

\( \implies V = \frac{3.75}{75} = 0.05\text{ V} \)

In simple words: This is a very low potential spot. Moving a charge here was very easy, so the "electrical pressure" at this point is low.

📝 Teacher's Note: This is another direct application of the potential formula. Remind students that potential is a scalar quantity—it has no direction.

🎯 Exam Tip: To divide 3.75 by 75, multiply both by 100 first: \( 375 / 7500 \). It makes the long division much clearer.

Practice Problems 5

Question 1. A work of 25 J and 30 J is done when 5 C charge is moved first to point A and then to point B from infinity. Calculate the potential difference between points A and B.

Answer:
\( W_A = 25\text{ J} ; W_B = 30\text{ J} \)
Work done to move the 5 C charge from A to B \( = W_B - W_A \)

\( \implies W = 30 - 25 = 5\text{ J} \)
Charge \( Q = 5\text{ C} \)
Potential difference between A and B

\( V = \frac{W}{Q} = \frac{5}{5} = 1\text{ V} \)
\( V = 1\text{ volt} \)
OR
\( W_A = 25\text{ J} \); \( W_B = 30\text{ J} \); charge \( = Q = 5\text{ C} \)
Potential at point A \( = V_1 = \frac{W_A}{Q} \)

\( \implies V_1 = \frac{25}{5} = 5\text{ volt} \)
Potential at point B \( = V_2 = \frac{W_B}{Q} \)

\( \implies V_2 = \frac{30}{5} = 6\text{ volt} \)
Potential difference A and B \( = V = V_2 - V_1 \)

\( \implies V = 6 - 5 = 1\text{ volt} \)

In simple words: Potential difference is just the "gap" between two spots. You can find it by calculating the work for that specific gap, or by finding the potentials for both spots and subtracting them.

📝 Teacher's Note: The second method (finding individual potentials) is usually safer for students as it clarifies what "potential" actually represents at each point.

🎯 Exam Tip: "Potential Difference" is also known as "Voltage." Use either term if they help you remember the concept.

Question 2. A charge of 25 C is moved from infinity to points A and B in an electric field when the work done to do so is 10 J and 10.5 J respectively. Calculate the potential difference between the points A and B.

Answer:
\( W_A = 10\text{ J} \)
Charge \( Q = 25\text{ C} \)
Potential at point A \( = V_1 = \frac{W_A}{Q} \)

\( \implies V_1 = \frac{10}{25} = 0.4\text{ V} \)
Also, \( W_B = 10.5\text{ J} \)
Potential at point B \( = V_2 = \frac{W_B}{Q} \)

\( \implies V_2 = \frac{10.5}{25} = 0.42\text{ volt} \)
Potential difference between A and B \( = V = V_2 - V_1 \)

\( \implies V = 0.42 - 0.40 \)

\( \implies V = 0.02\text{ volt} \)

In simple words: The two points are very close in their electrical pressure. The work needed to move between them is only the tiny difference between the work for A and the work for B.

📝 Teacher's Note: Reinforce subtraction with decimals. Many students might write 0.2 instead of 0.02.

🎯 Exam Tip: If points are labeled A and B, always assume B is the destination unless told otherwise, and subtract \( V_{\text{destination}} - V_{\text{source}} \).

Unit II

Exercise 2

(A) Objective Questions

Question 1. SI unit potential difference is :
(a) coulomb
(b) kelvin
(c) volt
(d) ampere

Answer: (c) volt
In simple words: We measure the "push" of a battery in Volts.

📝 Teacher's Note: Contrast all units: Coulomb (Charge), Kelvin (Temp), Volt (Potential), Ampere (Current).

🎯 Exam Tip: This is a fundamental factual question. Don't spend more than 10 seconds on it.

Question 2. Current in a circuit flows :
(a) in a direction from high potential to low potential
(b) in a direction from low potential to high potential
(c) in a direction of flow of electron
(d) in any direction

Answer: (a) in a direction from high potential to low potential
In simple words: Electricity is like water in a pipe—it naturally flows from a high-pressure spot to a low-pressure spot.

📝 Teacher's Note: This refers to "Conventional Current." Even though we know better now, we still use this rule for circuit analysis.

🎯 Exam Tip: "High to low" is the golden rule for current, just like "Hot to cold" is for heat.

Question 3. In a metallic conductor, electric current is thought to be due to movement of :
(a) ions
(b) amperes
(c) electrons
(d) protons

Answer: (c) electrons
In simple words: In metals like copper, the atoms are stuck in place, but they have tiny "loose" electrons that can zip through the wire.

📝 Teacher's Note: Remind students that protons are trapped in the nucleus and cannot move to carry current.

🎯 Exam Tip: The carriers in a metallic conductor are specifically "free electrons."

Question 4. Assuming that the charges of an electron is \( 1.6 \times 10^{-19}\text{ coulombs} \), the number of electrons passing through a section of wire per sec, when the wire carries a current of one ampere is :
(a) \( 0.625 \times 10^{19} \)
(b) \( 1.6 \times 10^{-19} \)
(c) \( 1.6 \times 10^{19} \)
(d) \( 0.627 \times 10^{-17} \)

Answer: (a) \( 0.625 \times 10^{19} \)
Explanation :
Charge \( = q = e = 1.6 \times 10^{-19}\text{ C} \)
Number of electrons \( = n = ? \)
Electric current \( = I = 1\text{ A} \)
Time \( = t = 1\text{ S} \)

\( I = \frac{q}{t} = \frac{ne}{t} \)

\( 1 = \frac{1.6 \times 10^{-19} \times n}{1} \)

\( \implies n = \frac{1}{1.6 \times 10^{-19}} = 0.625 \times 10^{19} \)

In simple words: One Ampere means one Coulomb passes every second. We just divide one Coulomb by the size of one electron to find how many electrons it takes.

📝 Teacher's Note: This is roughly the same constant as \( 6.25 \times 10^{18} \). Show how moving the decimal place changes the power from 18 to 19.

🎯 Exam Tip: This specific number is very common in MCQs. Memorize both formats: \( 6.25 \times 10^{18} \) or \( 0.625 \times 10^{19} \).

Question 5. Which of the following is best conductor of electricity?
(a) copper
(b) gold
(c) platinum
(d) silver

Answer: (d) silver
In simple words: Silver allows electricity to flow with the least amount of "traffic jam" or resistance.

📝 Teacher's Note: Mention that we use copper instead of silver in homes only because silver is too expensive, not because it's better.

🎯 Exam Tip: Silver is the #1 conductor, followed by Copper, then Gold.

(B) Subjective Questions

Question 1. What do you understand by the term electric cell?

Answer: Electric cell is an arrangement which maintains constant potential difference between conductors. A cell basically consists of two conducting rods, which are called electrodes, immersed in a solution, which is called the electrolyte.

In simple words: A cell is like a chemical factory in a bottle. It uses chemical reactions to constantly push electrons out of one rod and into another, creating the "pressure" (Voltage) needed for a circuit.

📝 Teacher's Note: Define "Potential difference" as the engine that drives current. Without the cell's PD, the electrons in the wire would just sit still.

🎯 Exam Tip: Use the two keywords: "electrodes" and "electrolyte" in your definition.

Question 2. Draw a neat and labelled diagram of simple voltaic cell showing clearly the direction of flow of conventional current and direction of flow of electrons.

Answer: Simple voltaic cell was invented by Alessandro Volta in year 1800. It was the first device which could create a constant potential difference between two plates with the help of chemical energy.
In simple words: The Zinc plate "shoots" electrons through the wire to the Copper plate. This flow of electrons is the electricity we use. By tradition, we draw the "current" arrow going the other way.

📝 Teacher's Note: The Zinc plate acts as the Anode (negative) and Copper as the Cathode (positive) in a Voltaic cell. Note that in a battery, electrons flow from negative to positive.

🎯 Exam Tip: Label the electrolyte and the specific metals (Zinc and Copper). Don't forget to mark the (+) and (-) signs clearly.

Question 3. Briefly describe the theory of simple voltaic cell.

Answer:
Theory of simple voltaic cell: Amongst the zinc and copper plates, zinc is more electro-positive (ionisation potential \( – 0.76\text{ V} \)) as compared to copper (ionisation potential \( + 0.34\text{ V} \)) in electrochemical series. The dilute sulphuric acid is used as an electrolyte in the ionised state.

\( \text{H}_2\text{SO}_4 \rightleftharpoons 2\text{H}^+ + \text{SO}_4^{2-} \)

When zinc plate comes in contact with (\( \text{H}^+ \)) hydrogen ions, it being more electro-positive, ionises to form zinc ions and free electrons.

\( \text{Zn} \longrightarrow \text{Zn}^{2+} + 2\text{e}^- \)

The free electrons so formed, take the passage of least resistance, and hence, move out in the external circuit through the copper wires. The zinc ions, however, enter in the dilute sulphuric acid. Since \( \text{Zn}^{2+} \) ions are positively charged, they repel \( \text{H}^+ \) ions in the acid solution, with the result that \( \text{H}^+ \) ions start crowding at the copper plate. The copper plate in turn loses its free electrons to \( \text{H}^+ \) ions, which form nascent hydrogen.

\( 2\text{H}^+ + 2\text{e}^- \longrightarrow 2\text{H} \)

The nascent hydrogen atoms so formed unite to form molecular hydrogen.

\( 2\text{H} \longrightarrow \text{H}_2 \)

From the above explanation, it is clear that free electrons actually drift from zinc to copper in external circuit, and hence, current should flow from zinc to copper. However, we still continue saying the electric charge flows from copper to zinc and this current is called conventional current, whereas the actual flow of electrons is from zinc to copper which constitutes electronic current. The emf between the zinc copper plate in the external circuit is \( 0.34\text{ V} - (-0.75\text{ V}) = 1.10\text{ V} \).

In simple words: The acid dissolves the zinc, releasing tiny electrons. These electrons run through the wire to get to the copper side. On the copper side, the acid's hydrogen joins the electrons to make bubbles of gas. This zipping of electrons is what we call electricity.

📝 Teacher's Note: The chemical reaction is the "pump" that creates the potential difference. The result is a steady 1.1 Volts (approx) until the zinc or acid is used up.

🎯 Exam Tip: Learn the chemical equations. They are mandatory for a descriptive answer on the theory of the cell.

Question 4. What do you understand by the following terms?
1. electric circuit
2. closed electric circuit
3. open electric circuit.

Answer:
1. Electric circuit : The path along which electric current flows is known as electric circuit.
2. Closed electric circuit : When the path of an electric circuit starting from one terminal of the cell, ends at the other terminal of cell, without any break, then such a circuit is called closed circuit.
3. Open electric circuit : When the path of an electric circuit, starting from one terminal of the cell, is broken at some point, then such a circuit is called open electric circuit.

In simple words: A circuit is a loop for electricity. If the loop is perfect (switch ON), it's "Closed." If there is a gap or a cut (switch OFF), it's "Open" and nothing flows.

📝 Teacher's Note: Use the "Bridge" analogy. A closed circuit is like a solid bridge where traffic flows; an open circuit is like a drawbridge that is raised—no cars can cross the gap.

🎯 Exam Tip: If a bulb doesn't glow, the first reason you should give is that it is an "open circuit."

Question 5. State two conditions necessary for a circuit, such that electric current flows through it.

Answer:
For the flow of electric current through a circuit, following are the necessary conditions :
1. Electric circuit must be closed or complete.
2. Every part of the circuit is a conductor.

In simple words: To get current, you need a full loop with no gaps, and everything in that loop must be made of metal or something else that allows electricity to pass.

📝 Teacher's Note: Remind students that even a tiny insulator like a piece of plastic or a speck of rust at a connection can stop the whole circuit from working.

🎯 Exam Tip: "Closed path" and "Presence of conductors" are the two fundamental requirements for any working circuit.

Question 6. Draw a neat diagram showing
1. closed electric circuit
2. open electric circuit.

Answer:
1. Closed circuit :
2. Open circuit :
In simple words: The first drawing shows the "ON" state with a bright bulb. The second drawing shows a "OFF" state because the wire is disconnected, and the bulb is dark.

📝 Teacher's Note: In diagrams, use a small circle with rays to show a glowing bulb, and a plain circle for a bulb that is off.

🎯 Exam Tip: When drawing an open circuit, clearly show the gap in the wire or the open position of the switch.

Question 7. Name four electric conductors and four electric insulators.

Answer:

• Conductors : Silver, copper, aluminium and iron.
• Insulators : Plastic, nylon, dry wood and rubber.

In simple words: Conductors are materials that let electricity "zip" through them easily (mostly metals). Insulators are materials that block electricity and keep it from going where it shouldn't.

📝 Teacher's Note: Remind students that water (especially tap water) is a conductor, which is why using electrical appliances near sinks is dangerous.

🎯 Exam Tip: If asked for a "best" conductor, the answer is always silver.

Question 8. (a) What do you understand by the term electric resistance?
(b) Why does the filament of an electric bulb in an electric circuit get white hot, but not the connecting wires?

Answer:
(a) Electric resistance : The obstruction offered to the passage of electric current by a material is called resistance of the material.
(b) Filament of an electric bulb is made up of tungsten having high resistance. Due to its high resistance, on passing electric current through it, an electrical energy changes into heat energy. So much heat is produced that filament of bulb becomes white hot and gives light. Resistance of connecting wires is very low and hence the connecting wires do not get heated.

In simple words: Resistance is like "friction" for electricity. The bulb's filament is like a very bumpy road that makes the electricity struggle so hard it gets hot enough to glow. The wires are like smooth highways where the electricity flows easily without getting hot.

📝 Teacher's Note: Use the analogy of rubbing your hands together—friction (resistance) creates heat. Tungsten is chosen for filaments specifically because it has a high melting point and high resistance.

🎯 Exam Tip: The keyword here is "Tungsten." Mentioning its "high resistance" and "high melting point" is essential for full marks.

Question 9. Is it correct to say that a resistance wire is an insulator or a bad conductor? Explain your answer.

Answer: It is not correct to say that a resistance wire is an insulator or bad conductor. From resistance, it is implied that a given material will conduct electricity, but will also offer obstruction to the passage of electric current.

In simple words: No, a resistance wire is like a "heavy door"—it's hard to push open, but you can still go through. An insulator is like a "brick wall"—it doesn't let anyone through at all.

📝 Teacher's Note: Distinguish between the three: Conductors (low resistance), Resistance Wires (controlled/high resistance), and Insulators (extremely high/infinite resistance).

🎯 Exam Tip: Use the comparison that an insulator has "infinite resistance," while a resistance wire has a "finite but high resistance."

Question 10. (a) What do you understand by the term series circuit?
(b) State two characteristics of resistances in the series circuit.
(c) Draw a diagram showing two bulbs connected in series to a dry cell.

Answer:
(a) Series circuit : When a number of resistances are connected in an electrical circuit in such a way that positive of one resistance acts as the negative of the other resistance, then resistances are said to be in series.
OR
A number of resistors are said to be in series if these are joined end to end and same current flows through each one of them when a potential difference is applied across the combination.
(b) Characteristics of resistances in series :
1. The sum total of resistances in series increases with increase in number of resistors.
2. The potential difference remaining constant, the current in series circuit decreases with the increase in number of resistors in series.
3. All the elements in series circuit work simultaneously. If the circuit is broken anywhere between the elements, none of the elements work.
(c)
In simple words: A series circuit is like a "single-track road." Every car (electron) must pass through every toll booth (bulb) one after the other. If one booth closes, the whole road is blocked.

📝 Teacher's Note: Highlight the most famous problem with series circuits: if one bulb fuses, the entire string (like old Christmas lights) goes dark.

🎯 Exam Tip: Remember: In series, Current is the same in all bulbs, but Voltage is divided among them.

Question 11. (a) What do you understand by the term parallel circuit?
(b) State two characteristics of resistance in the parallel circuit.
(c) Draw a diagram showing two bulbs connected in parallel to a dry cell.

Answer:
(a) Parallel circuit : When a number of resistances (bulbs) are connected in an electrical circuit in such a way that all of them are connected to common positive and common negative terminal of a cell, then the resistance (bulbs) are said to be connected in parallel.
OR
A number of resistors are said to be connected in parallel if one end of each resistor is connected to one point and other end of each resistor is connected to another point so that the potential difference across each resistor is same.
(b) Characteristics of resistances in Parallel :
1. The sum total of resistances in parallel decreases with the increase in number of resistors.
2. The current flowing in any resistor in parallel will be inversely proportional to resistance i.e., more the resistance, less the current.
3. Each resistor in parallel functions independently with respect to the other resistors in parallel.
(c)
In simple words: A parallel circuit is like a "multi-lane highway." Electricity can choose which lane to take. If one bulb breaks, the electricity just keeps flowing through the other lanes, so the rest stay on.

📝 Teacher's Note: This is how houses are wired. If your bedroom light goes out, your kitchen light stays on because they are in parallel.

🎯 Exam Tip: Remember: In parallel, Voltage is the same across all bulbs, but Current is divided between them.