Current Electricity Test

Result

Current Electricity Test - 97

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Weekly Quiz Competition

Question 1
4 / -1

In the circuit shown, the current through 8 ohm is same before and after connecting E. The value of E is

A
12 V

B
6 V

C
4 V

D
2 V

Solution
Question 2
4 / -1

If power dissipated in the 9 Ω resistor in the circuit shown is 36 Watt, the potential difference across the 2 Ω resistor is

A
2 volt

B
4 volt

C
8 volt

D
10 volt

Solution
Question 3
4 / -1

The magnitude and direction of the current in the circuit will be

A
7/3 A from a to b through e

B
7/3 A from b to a through e

C
1A from b to a through e

D
1A from a to b through e

Solution
Question 4
4 / -1

By a cell a current of 0.9 A flows through 2 ohm resistor and 0.3 A through 1 ohm resistor. The internal resistance of the cell is

A
0.5 Ω

B
1.0 Ω

C
1.2 Ω

D
2.0 Ω

Solution
Question 5
4 / -1

A student measures the terminal potential difference(V) of a cell (of e.m.f E and internal resistance r) as a function of the current (I) flowing through it. The slope, and intercept, of the graph between V and I,then, respectively, equal

A
E and –r

B
–r and E

C
r and –E

D
–E and r

Solution
Question 6
4 / -1

A cell of e.m.f. E is connected with an external resistance R, then p.d. across cell is V. The internal resistance of cell will be

A

B

C

D

Solution
Question 7
4 / -1

Two cells, each of e.m.f. E and internal resistance r are connected in parallel between the resistance R. The maximum energy given to the resistor will be, only when

A
R = r/2

B
R = r

C
R = 2r

D
R = 0

Solution
Question 8
4 / -1

Kirchhoff s first law i.e. ∑i = 0 at a junction is based on the law of conservation of

A
Charge

B
Energy

C
Momentum

D
Angular momentum
Question 9
4 / -1

The figure below shows currents in a part of electric circuit. The current is p

A
1.7 amp

B
3.7 amp

C
1.3 amp

D
1 amp

Solution
Question 10
4 / -1

See the electrical circuit shown in this figure. Which of the following equations is a correct equation for it?

A
ε₁– (i₁ + i₂) R –i₁ r₁ = 0

B
ε₂ – i₁r₂ – ε₁ – i₁r₁ = 0

C
–ε₂ – (i_i + i₂) R – i₂r₂ = 0

D
ε₁ – (i₁ + i₂) R + i₁r₁ = 0