Current Electricity Easy Questions and Answers | Page - 15

Current Electricity

n equal resistors are first connected in series and then connected in parallel. What is the ratio of the maximum to the minimum resistance ?

1. n
1. 1/n²
1. n²
1. 1/ n

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In series, R_s = nR
In parallel, 1 = 1 + 1 + ... n terms
R_p R R

∴ R_s/R_p = n² /1 = n²

Correct Option: C

In series, R_s = nR
In parallel, 1 = 1 + 1 + ... n terms
R_p R R

∴ R_s/R_p = n² /1 = n²

In the network shown in the Fig, each resistance is 1Ω. The effective resistance between A and B is

1. 4/3 Ω
1. 3/2 Ω
1. 7 Ω
1. 8/7 Ω

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At A current is distributed and at B currents are collected. Between A and B, the distribution is symmetrical. It has been shown in the figure. It appears that current in AO and OB remains same. At O, current i₄ returns back without any change. If we detach O from AB there will not be any change in distribution. Now, CO & OD will be in series hence its total resistance = 2Ω
It is in parallel with CD, so, equivalent resistance = 2 x 1 = 2 Ω
2 + 1 3

This equivalent resistance is in series with AC & DB, so, total resistance
= 2 + 1 + = 8 Ω
3 3

Now 8/3 Ω is parallel to AB, that is, 2Ω, so total resistance

ABCD = 1 + 2 + 1 = 8 Ω
3 3

Correct Option: D

At A current is distributed and at B currents are collected. Between A and B, the distribution is symmetrical. It has been shown in the figure. It appears that current in AO and OB remains same. At O, current i₄ returns back without any change. If we detach O from AB there will not be any change in distribution. Now, CO & OD will be in series hence its total resistance = 2Ω
It is in parallel with CD, so, equivalent resistance = 2 x 1 = 2 Ω
2 + 1 3

This equivalent resistance is in series with AC & DB, so, total resistance
= 2 + 1 + = 8 Ω
3 3

Now 8/3 Ω is parallel to AB, that is, 2Ω, so total resistance

ABCD = 1 + 2 + 1 = 8 Ω
3 3

You are given several identical resistances each of value R = 10Ω and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistances of 5Ω which can carry a current of 4 ampere. The minimum number of resistances of the type R that will be required for this job is

1. 4
1. 10
1. 8
1. 20

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To carry a current of 4 ampere, we need four paths, each carrying a current of one ampere. Let r be the resistance of each path. These are connected in parallel. Hence, their equivalent resistance will be r/4. According to the given problem
r = 5 or r = 20 Ω
4

For this propose two resistances should be connected. There are four such combinations. Hence, the total number of resistance = 4 × 2 = 8.

Correct Option: C

To carry a current of 4 ampere, we need four paths, each carrying a current of one ampere. Let r be the resistance of each path. These are connected in parallel. Hence, their equivalent resistance will be r/4. According to the given problem
r = 5 or r = 20 Ω
4

For this propose two resistances should be connected. There are four such combinations. Hence, the total number of resistance = 4 × 2 = 8.

Current through 3 Ω resistor is
0.8 amp., then potential drop through 4Ω resistor is

1. 9.6 V
1. 2.6 V
1. 4.8 V
1. 1.2 V

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Voltage across 3 Ω resistance = 3 × 0.8 = 2.4V
This voltage is the same across 6 Ω resistance.
Hence current through this resistance
i = V = 2.4 = 0.4 amp
R 6

Total current in the circuit = 0.8 + 0.4 = 1.2 amp
Voltage across 4 Ω resistance = 4 × 1.2 = 4.8 volts

Correct Option: C

Voltage across 3 Ω resistance = 3 × 0.8 = 2.4V
This voltage is the same across 6 Ω resistance.
Hence current through this resistance
i = V = 2.4 = 0.4 amp
R 6

Total current in the circuit = 0.8 + 0.4 = 1.2 amp
Voltage across 4 Ω resistance = 4 × 1.2 = 4.8 volts

Three resistances each of 4 Ω are connected to form a triangle. The resistance between any two terminals is

1. 12 Ω
1. 2 Ω
1. 6Ω
1. 8/3Ω

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The two resistances are connected in series and the resultant is connected in parallel with the third resistance.
∴ R' = 4Ω + 4Ω = 8Ω and 1 = 1 + 1 = 3
R" 8 4 8

Correct Option: D

The two resistances are connected in series and the resultant is connected in parallel with the third resistance.
∴ R' = 4Ω + 4Ω = 8Ω and 1 = 1 + 1 = 3
R" 8 4 8