Current Electricity Test

Result

Current Electricity Test - 24

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Weekly Quiz Competition

Question 1
4 / -1

In the following circuit reading of voltmeter V is

A
12 V

B
8 V

C
20 V

D
16 V

Solution
Question 2
4 / -1

Four cells, each of emf E and internal resistance r, are connected in series across an external resistance R. By mistake one of the cells is connected in reverse. Then, the current in the external circuit is

A

B

C

D

Solution
Question 3
4 / -1

A circuit consists of three batteries of emf E₁ =1V, E₂ = 2 V and E₃ = 3 V and internal resistances 1 Ω, 2 Ω and 1 Ω respectively which are connected in parallel as shown in the figure. The potential difference between points P and Q is

A
1.0 V

B
2.0 V

C
2.2 V

D
3.0 V

Solution
Question 4
4 / -1

Two identical cells whether connected in parallel or in series gives the same current, when connected to an external resistance 1.5 Ω. Find the value of internal resistance of each cell

A
1 Ω

B
0.5 Ω

C
zero

D
1.5 Ω

Solution
Question 5
4 / -1

For the resistance network shown in the figure, choose the correct option(s).

A
The current through PQ is zero

B
I₁ = 3A

C
The potential as S is less than that at Q

D
I₂ = 2A

Solution
Question 6
4 / -1

In the circuit shown, the galvanometer G of resistance 60 Ω is shunted by a resistance r = 0.02 Ω. The current through R is nearly 1 A. The value of resistance R is (in ohms) is nearly

A
50 Ω

B
100 Ω

C
200 Ω

D
400 Ω

Solution
Question 7
4 / -1

A
I, II and III

B
II and III

C
Only II

D
Only I

Solution
Question 8
4 / -1

The maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal resistance r, will be

A

B

C

D

Solution
Question 9
4 / -1

For the circuit shown in figure the potential difference between A and B will be (in volt)

A
2

B
1.5

C
1.0

D
zero

Solution
Question 10
4 / -1

In the circuit given here, the points A, B and C are 70 V, 0V, 10 V respectively. Then,

A
the point D will be at a potential of 60 V

B
the point D will be at a potential of 20 V

C
currents in the paths AD, DB and DC are in the ratio of 1 : 2 : 3

D
currents in the paths AD, DB and DC are in the ratio of 3 : 2 : 1

Solution