Current Electricity Test

Result

Current Electricity Test - 100

Score
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Rank
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TIME Taken - -

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Weekly Quiz Competition

Question 1
4 / -1

The current in the arm CD of the circuit will be

A
i₁ + i₂

B
i₂+ i₃

C
i₁ + i₃

D
i₁–i₂ + i₃

Solution
Question 2
4 / -1

When a resistance of 2 ohm is connected across the terminals of a cell, the current is 0.5 A. When the resistance is increased to 5 ohm, the current is 0.25 A. The e.m.f. of the cell is

A
1.0 V

B
1.5 V

C
2.0 V

D
2.5 V

Solution
Question 3
4 / -1

4 cells each of e.m.f 2 V and internal resistance of 1 Ω are connected in parallel to a load resistor of 2 Ω. Then thecurrent through the load resistor is

A
2 A

B
1.5 A

C
1 A

D
0.888 A

Solution
Question 4
4 / -1

If V_AB = 4V in the given figure, then resistance X will be

A
5 Ω

B
10 Ω

C
15Ω

D
20 Ω

Solution
Question 5
4 / -1

A storage battery has e.m.f. 15 volt and internal resistance 0.05 ohm. Its terminal voltage when it is delivering 10 ampere is

A
30 volt

B
1.00 volt

C
14.5 volt

D
15.5 volt

Solution
Question 6
4 / -1

The internal resistance of a cell of e.m.f. 12 V is 5 × 10^–2Ω. It is connected across an unknown resistance. Voltage across the cell, when a current of 60 A is drawn from it, is

A
15 V

B
12 V

C
9 V

D
6 V

Solution
Question 7
4 / -1

The current in the given circuit is

A
0.1 A

B
0.2 A

C
0.3 A

D
0.4 A

Solution
Question 8
4 / -1

When a resistor of 11 Ω is connected in series with an electric cell, the current flowing in it is 0.5 A. Instead, when a resistor of 5 Ω is connected to the same electric cell in series, the current increases by 0.4 A. The internal resistance of the cell is

A
1.5 Ω

B
2 Ω

C
2.5 Ω

D
3.5 Ω

Solution
Question 9
4 / -1

The internal resistance of a cell is the resistance of

A
Electrodes of the cell

B
Vessel of the cell

C
Electrolyte used in the cell

D
Material used in the cell
Question 10
4 / -1

How much work is required to carry a 6μC charge from the negative terminal to the positive terminal of a 9 V battery?

A
54 × 10^–3 J

B
54 × 10^–6 J

C
54 × 10^–9 J

D
54 × 10^–12 J

Solution