CBSE Class 10 Science Electricity and Magnetic Effects of Electric Current Assignment Set 09

Answer: (a) 1 volt is the potential difference between two points when 1 joule of work is done to move 1 coulomb of charge between them.
(b) 1 kWh is the energy used when a device of power 1 kW works for 1 hour; 1 kWh = 3.6 × 10⁶ J.
(c) 1 ohm is the resistance of a conductor that allows a current of 1 ampere when the potential difference across it is 1 volt.

Teacher's Note:
1. Each unit is defined from its own formula: V = W/Q, E = P × t and R = V/I.
2. Do not forget 1 kWh = 3.6 × 10⁶ J.

Answer: Ohm's law: at constant temperature the current through a conductor is directly proportional to the potential difference across its ends, so V = IR.
Joule's law: the heat produced in a resistor is directly proportional to the square of the current, to the resistance and to the time for which current flows: H = I²Rt.

Teacher's Note:
1. Ohm's law is valid only when the temperature stays constant.
2. Joule's law explains heaters and fuses.

Answer: Resistance is the property of a conductor that opposes the flow of charge through it; R = V/I and its unit is the ohm.
It depends on (i) the length of the conductor (R ∝ l), (ii) its area of cross-section (R ∝ 1/A), (iii) the material (resistivity ρ) and (iv) the temperature. So R = ρl/A.

Teacher's Note:
1. Resistance rises with length and falls with area.
2. Metals usually have higher resistance at higher temperature.

Answer: In parallel the potential difference V across each resistor is the same, and the total current is I = I₁ + I₂ + I₃.
Using Ohm's law, I = V/R₁ + V/R₂ + V/R₃ = V (1/R₁ + 1/R₂ + 1/R₃). If R_p is the equivalent resistance, I = V/R_p, so 1/R_p = 1/R₁ + 1/R₂ + 1/R₃.

Teacher's Note:
1. The same potential difference acts across each branch.
2. The equivalent resistance is smaller than the smallest branch.

Answer: A solenoid is a long coil of insulated wire wound closely in the form of a cylinder. Its field lines look like those of a bar magnet: they run parallel and equally spaced inside it (a uniform field) and spread out from one end and return at the other end outside.
The field depends on (i) the current in the coil, (ii) the number of turns per unit length and (iii) the core material; a soft iron core greatly strengthens it.

Teacher's Note:
1. Parallel straight lines inside show the field is uniform.
2. One end of a solenoid acts as the north pole and the other as the south pole.

Answer: Properties: 1. They emerge from the north pole and enter the south pole outside the magnet, and form closed curves. 2. The tangent at any point gives the direction of the field there. 3. They are closer where the field is stronger. 4. They never intersect.
If two lines crossed, a compass needle at the crossing point would have to point in two directions at once, which is impossible. Hence they cannot cross.

Teacher's Note:
1. The direction of the field at a point is unique.
2. Crowded lines mean a stronger field.

Answer: Imagine holding the wire in your right hand with the thumb pointing in the direction of the current. The fingers then curl around the wire in the direction of the magnetic field lines, which are concentric circles around the wire.

Teacher's Note:
1. The thumb shows the current and the curled fingers show the field.
2. Reversing the current reverses the field direction.

Answer: Stretch the thumb, forefinger and middle finger of the left hand mutually perpendicular. If the forefinger points along the magnetic field and the middle finger along the current, the thumb gives the direction of the force on the conductor.
It is used to find the direction of the force on a current-carrying conductor placed in a magnetic field.

Teacher's Note:
1. Forefinger is the field, middle finger is the current and thumb is the force.
2. The force is largest when the current is perpendicular to the field.

Answer: In parallel each appliance gets the full supply voltage and can be switched on or off independently. In series the same current flows through all, the voltage is divided, and if one appliance fails the whole circuit stops.

Teacher's Note:
1. Parallel connection keeps the voltage the same for every device.
2. It also means total resistance is low enough to draw the current needed.

Answer: I = V/R = 12 / 20 = 0.6 A.

Teacher's Note:
1. Apply Ohm's law directly.
2. Write the unit ampere in the answer.

Answer: 1/R = 1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = 1, so R = 1 Ω.

Teacher's Note:
1. Add the reciprocals, then invert.
2. The answer is less than 2 Ω, the smallest resistor.

Answer: (a) Total resistance = 4 + 6 = 10 Ω, so I = 20/10 = 2 A.
(b) V across 6 Ω = IR = 2 × 6 = 12 V.

Teacher's Note:
1. The same current flows through resistors in series.
2. The voltages across the two resistors add up to 20 V.

Answer: The volume stays constant, so doubling the length halves the area of cross-section. R = ρl/A becomes ρ(2l)/(A/2) = 4ρl/A.
New resistance = 4 × 10 = 40 Ω.

Teacher's Note:
1. Length doubles and area halves, so resistance becomes four times.
2. Volume of the wire stays the same on stretching.

Answer: I = P/V = 60/220 ≈ 0.27 A.
R = V²/P = (220)²/60 = 48400/60 ≈ 807 Ω.

Teacher's Note:
1. Use P = VI and R = V²/P.
2. The hot filament resistance is much higher than when cold.

Answer: Energy = power × time = 2 kW × 5 h = 10 kWh (10 units).
Cost = 10 × ₹6 = ₹60.

Teacher's Note:
1. One unit of electrical energy is 1 kWh.
2. Cost equals units consumed times the rate per unit.

Answer: H = I²Rt = 5² × 20 × 120 = 25 × 20 × 120 = 60 000 J (60 kJ).

Teacher's Note:
1. Convert the time into seconds before substituting.
2. Joule's law gives the heat in joules.