CBSE Class 10 Physics Electricity MCQs Set 12

Answer: (a) series circuit

Explanation:
1. In a series circuit the resistors are joined end to end, so the same current flows through each resistor.
2. In a parallel circuit the current divides among the branches.

Teacher's Note:
1. Series: same current, different potential difference.
2. Parallel: same potential difference, different currents.

Answer: (b) potential difference

Explanation:
1. In a parallel combination all the resistors are connected between the same two points.
2. So the potential difference across each resistor is the same, while the currents are different.

Teacher's Note:
1. The current from the source splits among the parallel branches.
2. This is why household appliances are connected in parallel.

Answer: (a) 1/2

Explanation:
1. By Ohm’s law I = V/R.
2. At constant V, if R becomes 2R, then I′ = V/2R = I/2.

Teacher's Note:
1. Current is inversely proportional to resistance at constant voltage.
2. Doubling R halves the current.

Answer: (b) V = V₁ + V₂ + V₃

Explanation:
1. In a series combination the same current flows through all the resistors.
2. The total voltage of the battery is divided among them, so V = V₁ + V₂ + V₃.

Teacher's Note:
1. In series, voltages add and the current is the same.
2. In parallel, currents add and the voltage is the same.

Answer: (b) > R

Explanation:
1. In series R_S = R + R + R = 3R.
2. 3R is greater than R, so the equivalent resistance is > R.

Teacher's Note:
1. A series combination always gives an equivalent resistance greater than the largest resistance.
2. A parallel combination gives a value less than the smallest resistance.

Answer: (b) the same current flowing through them when connected in series

Explanation:
1. In series the same current flows through every resistor, and the potential differences are different.
2. In parallel the potential difference across each resistor is the same, and the currents are different.
3. So only statement (b) is correct.

Teacher's Note:
1. Series: same current. Parallel: same potential difference.
2. Remember this pair of rules to answer many circuit questions quickly.

Answer: (c) voltage-drop is same across both

Explanation:
1. In a parallel combination the potential difference (voltage drop) is the same across each resistor.
2. The current is larger in the smaller resistance (I = V/R).

Teacher's Note:
1. Same voltage means current is inversely proportional to resistance.
2. The statement that current is the same applies to a series combination.

Answer: (a) R₁R₂/(R₁ + R₂)

Explanation:
1. For parallel resistors 1/R = 1/R₁ + 1/R₂ = (R₁ + R₂)/R₁R₂.
2. So R = R₁R₂/(R₁ + R₂).

Teacher's Note:
1. For two resistors in parallel the equivalent resistance is product over sum.
2. It is always smaller than the smaller resistance.

Answer: (a) 1 Ω

Explanation:
1. For parallel resistors 1/R = 1/3 + 1/3 + 1/3 = 1.
2. So R = 1 Ω.

Teacher's Note:
1. n equal resistors r in parallel give r/n.
2. Here 3/3 = 1 Ω.

Answer: (c) 50 Ω

Explanation:
1. In series the equivalent resistance is the sum of the individual resistances.
2. X = 10 + 40 = 50 Ω.

Teacher's Note:
1. Series: R = R₁ + R₂.
2. The result is greater than the larger resistance.

Answer: (b) 1/25 Ω

Explanation:
1. The minimum resistance is obtained when all the resistors are connected in parallel.
2. For n equal resistors r in parallel, R = r/n = (1/5)/5 = 1/25 Ω.

Teacher's Note:
1. Parallel combination gives a resistance smaller than the smallest individual resistance.
2. Series would give the maximum value, 5 × 1/5 = 1 Ω.

Answer: (a) 1 Ω

Explanation:
1. The maximum resistance is obtained by connecting the resistors in series.
2. R = 5 × (1/5) = 1 Ω.

Teacher's Note:
1. Series combination gives the maximum resistance.
2. The least resistance of the same five resistors in parallel would be 1/25 Ω.

Answer: (c) 1/100 Ω

Explanation:
1. The lowest resistance is obtained when all the resistors are connected in parallel.
2. R = (1/10)/10 = 1/100 Ω.

Teacher's Note:
1. Parallel connection of n equal resistors r gives r/n.
2. Series would give 10 × 1/10 = 1 Ω.

Answer: (a) 2 Ω

Explanation:
1. Maximum resistance is obtained when all the resistors are connected in series.
2. R = 1/2 + 1/2 + 1/2 + 1/2 = 2 Ω.

Teacher's Note:
1. Series combination gives the largest resistance and parallel combination gives the smallest.
2. The equivalent resistance in series is the sum of the individual resistances.

Answer: (d) 16

Explanation:
1. In series R_S = 4 × 8 = 32 Ω.
2. In parallel R_P = 8/4 = 2 Ω.
3. Ratio R_S/R_P = 32/2 = 16.

Teacher's Note:
1. For n identical resistors the ratio R_S/R_P is n².
2. Here n = 4, so the ratio is 16.

Answer: (d) 4/3 Ω

Explanation:
1. Between two vertices, one side (2 Ω) is in parallel with the other two sides in series (2 + 2 = 4 Ω).
2. R = (2 × 4)/(2 + 4) = 8/6 = 4/3 Ω.

Teacher's Note:
1. Identify which resistors are in series and which are in parallel before calculating.
2. For a triangle of three equal resistors r, the resistance between two corners is 2r/3.

Answer: (a) 20 Ω

Explanation:
1. Each piece has a resistance of 20/4 = 5 Ω.
2. The four pieces in series give 5 + 5 + 5 + 5 = 20 Ω.

Teacher's Note:
1. Joining the pieces in series gives the original wire back.
2. Resistance of the whole wire is unchanged: 20 Ω.

Answer: (a) in series

Explanation:
1. In a series combination the total potential difference is divided among the resistors, V = V₁ + V₂ + V₃.
2. So a high potential is distributed by connecting resistors in series.

Teacher's Note:
1. In parallel every resistor gets the full potential difference, but the current is shared.
2. A potential divider uses resistors in series.

Answer: (a) 2 ohm to 50 ohm

Explanation:
1. The minimum resistance is obtained when all five are connected in parallel: 10/5 = 2 Ω.
2. The maximum resistance is obtained when all five are connected in series: 5 × 10 = 50 Ω.
3. So the value lies between 2 Ω and 50 Ω.

Teacher's Note:
1. Series gives the maximum and parallel gives the minimum resistance.
2. Any other combination gives a value between these two limits.

Answer: (b) 2R

Explanation:
1. Each group of two resistors R in parallel has R/2.
2. Four such groups in series give 4 × R/2 = 2R.

Teacher's Note:
1. First reduce each parallel group, then add the groups in series.
2. 2 resistors in parallel give half the resistance.

Answer: (b) 3

Explanation:
1. The required resistance (2 Ω) is less than 6 Ω, so the resistors must be in parallel.
2. For n equal resistors R_P = 6/n = 2, so n = 3.

Teacher's Note:
1. Parallel combination decreases the resistance.
2. n resistors of 6 Ω in parallel give 6/n.

Answer: (b) 6 Ω and 4 Ω

Explanation:
1. Let the resistors be R₁ and R₂: R₁ + R₂ = 10 and R₁R₂/(R₁ + R₂) = 2.4, so R₁R₂ = 24.
2. The numbers with sum 10 and product 24 are 6 and 4.

Teacher's Note:
1. Use sum = 10 and product = 24 to find the two numbers.
2. Check: 6 × 4/10 = 2.4 Ω.

Answer: (c) < 1 Ω

Explanation:
1. The minimum resistance is obtained when all the resistors are connected in parallel.
2. The parallel combination is always less than the smallest resistance, which is 1 Ω.
3. So the minimum resistance is < 1 Ω.

Teacher's Note:
1. Parallel resistance is smaller than the smallest individual resistance.
2. Here 1/R = 1/2 + 1/4 + 1 + 1/100, so R ≈ 0.5 Ω.

Answer: (c) 3, 3, 3 ohm

Explanation:
1. All the options add up to 9 Ω, so the series condition does not decide.
2. Check the parallel condition: (4, 4, 1) gives 0.67 Ω, (6, 2, 1) gives 0.6 Ω, (3, 3, 3) gives 1 Ω and (2, 3, 4) gives 0.92 Ω.
3. Only 3 Ω, 3 Ω, 3 Ω gives an effective resistance of 1 Ω in parallel.

Teacher's Note:
1. Use both conditions together to find the right option.
2. For three equal resistors R, R_P = R/3 and R_S = 3R.

Answer: (a) 4 Ω, 12 Ω

Explanation:
1. If the coils are 4 Ω and 12 Ω, they give 4 Ω and 12 Ω separately.
2. In series: 4 + 12 = 16 Ω. In parallel: (4 × 12)/(4 + 12) = 3 Ω.
3. All four values 3, 4, 12 and 16 are obtained.

Teacher's Note:
1. Test each option for series and parallel results.
2. The other options give values like 16, 3.9, 3.4 and 3.75 in parallel.

Answer: (d) R < r₃

Explanation:
1. In a parallel combination 1/R = 1/r₁ + 1/r₂ + 1/r₃, which is greater than 1/r₃.
2. So R is less than the smallest resistance r₃.

Teacher's Note:
1. The equivalent resistance in parallel is always less than the least individual resistance.
2. Adding more parallel branches lowers the resistance.

Answer: (c) potential difference across larger resistance is more

Explanation:
1. In series the same current flows through both resistances, so (b) is wrong.
2. The potential difference V = IR is larger across the larger resistance.

Teacher's Note:
1. Series: same current. The voltage divides in proportion to the resistances.
2. Parallel: same voltage, the current divides.

Answer: (b) parallel

Explanation:
1. In a parallel circuit every appliance gets the same voltage, and each can be switched on or off independently.
2. Household circuits are therefore connected in parallel.

Teacher's Note:
1. If one appliance fails, the others continue to work.
2. Appliances of different ratings need different currents, which parallel wiring gives.

Answer: (c) (x + y)²/xy

Explanation:
1. In series Z = x + y.
2. In parallel M = xy/(x + y).
3. Z/M = (x + y)/(xy/(x + y)) = (x + y)²/xy.

Teacher's Note:
1. Write both equivalent resistances first and then divide.
2. For equal resistors Z : M = 4 : 1.

Answer: (a) 2 V

Explanation:
1. V = IR = 0.2 × 10 = 2 V across the 10 Ω resistor.
2. In a parallel combination the voltage across each resistor equals the battery voltage, so the battery supplies 2 V.

Teacher's Note:
1. Use Ohm’s law on the resistor whose current is known.
2. The same voltage appears across the other resistor.

Answer: (a) are connected in parallel

Explanation:
1. All the devices get the same voltage of 220 V.
2. The potential difference is the same only for devices connected in parallel.

Teacher's Note:
1. In a parallel connection each device works independently.
2. Currents and resistances of different devices are different.

Answer: (d) 10 V

Explanation:
1. In a parallel combination the potential difference across every resistor is the same as the battery voltage.
2. So the potential difference across the 3 Ω resistor is 10 V.

Teacher's Note:
1. Resistors in parallel share the same two end points.
2. Their currents differ, but the voltage is the same.

Answer: (b) n²X

Explanation:
1. In parallel X = R/n, so R = nX.
2. In series the equivalent resistance is nR = n × nX = n²X.

Teacher's Note:
1. Express R in terms of X first.
2. Series gives nR and parallel gives R/n, so their ratio is n².

Answer: (a) 40 Ω

Explanation:
1. For the same length and metal R ∝ 1/A.
2. The thicker wire has 3 times the area, so its resistance is 10 Ω and the thinner wire has 3 × 10 = 30 Ω.
3. In series the total is 10 + 30 = 40 Ω.

Teacher's Note:
1. A larger area gives a smaller resistance.
2. The thinner wire has three times the resistance of the thicker wire.

Answer: (a) P-(iii), Q-(ii), R-(iv), S-(i)

Explanation:
1. The current divides at a junction in a parallel circuit, so P matches (iii).
2. Two or more cells connected together form a battery, so Q matches (ii).
3. The current is the same at all points in a series circuit, so R matches (iv).
4. A fuse protects electrical appliances, so S matches (i).

Teacher's Note:
1. Remember: series has a common current, parallel has a common voltage.
2. A fuse protects against excess current.

Answer: (d) A is false, but R is true.

Explanation:
1. A is false: in series the equivalent resistance is the sum of the resistances, so it increases the resistance and decreases the current.
2. R is true: in parallel 1/R_P = 1/R₁ + 1/R₂ + ..., so R_P is smaller than the smallest resistance.
3. So A is false but R is true.

Teacher's Note:
1. To decrease the resistance or increase the current, resistors are joined in parallel.
2. Series combination increases resistance.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is true: end-to-end connection is a series combination.
2. R is true: in parallel, the equivalent resistance is less than the smallest resistance.
3. But R does not explain the definition of series connection, so R is not the correct explanation of A.

Teacher's Note:
1. Series gives larger resistance, parallel gives smaller resistance.
2. Both statements are correct but they talk about different things.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A is true: in a parallel circuit the current divides among the gadgets, so each gets the current it needs, and household circuits are parallel.
2. R is true: the total resistance decreases in a parallel circuit, which helps appliances of different resistances to work together.
3. R supports A, so R is the correct explanation.

Teacher's Note:
1. In parallel circuits each appliance can be switched on or off independently.
2. Appliances of different ratings need different currents, which parallel wiring allows.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is true: the small bulbs of decorative fairy lights are connected in series.
2. R is true: in series, if one component fails the whole circuit is broken.
3. But R is a drawback of series connection and does not explain why fairy lights are connected in series.

Teacher's Note:
1. In series the same current flows through all the bulbs and each gets a small part of the supply voltage.
2. That is why a fused bulb makes the whole string go off.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is true: in a series circuit there is only one path, so the same current flows through every part.
2. R is also true: different devices need different currents (depending on their power ratings).
3. But R does not explain why the current is constant in a series circuit, so R is not the correct explanation.

Teacher's Note:
1. That is why series circuits are not used for homes, where appliances need different currents.
2. Both statements are facts, but they are not cause and effect.