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Current Electricity MSQ

10 Questions MCQ Test Topic wise Tests for IIT JAM Physics | Current Electricity MSQ

Description

This mock test of Current Electricity MSQ for Physics helps you for every Physics entrance exam. This contains 10 Multiple Choice Questions for Physics Current Electricity MSQ (mcq) to study with solutions a complete question bank. The solved questions answers in this Current Electricity MSQ quiz give you a good mix of easy questions and tough questions. Physics students definitely take this Current Electricity MSQ exercise for a better result in the exam. You can find other Current Electricity MSQ extra questions, long questions & short questions for Physics on EduRev as well by searching above.

*Multiple options can be correct

QUESTION: 1

A battery of internal resistance 2Ω is connected to a variable resistor whose value can vary from 4Ω to 10Ω. The resistance is initially set at 4Ω. If the resistance is now increased then, which of the statement is not correct.

A.
power consumed by it may increase or may decrease
B.
power consumed by it will decrease
C.
power consumed will first increase then decrease.
D.
power consumed by it will increase

Solution:

Power maximum when r = R.
So, power consumed by it will decrease. for R > r.

The correct answers are: power consumed by it will increase, power consumed by it may increase or may decrease, power consumed will first increase then decrease.

*Multiple options can be correct

QUESTION: 2

In the figure a conductor of non-uniform cross-section is shown. A steady current I flows in it.

A.
The electric field at B is more than at A.
B.
The thermal power generated at B is more than at A in an element of small width.
C.
The electric field at A is more than at B.
D.
The thermal power generated at A is more than at B in an element of small width.

Solution:

The area of cross-section of conductor at point A is less than that at point B. So current density at A is higher. Hence, the electric field at A is more than at B and the thermal power generated at A is more than at B in an element of small same width, since resistance at A is greater.

The correct answers are: The electric field at A is more than at B., The thermal power generated at A is more than at B in an element of small width.

*Multiple options can be correct

QUESTION: 3

Two cells of unequal emfs E₁ and E₂ and internal resistances r₁ and r₂ are joined as shown in figure. V_P and V_Q are the potential at P and Q respectively.

A.
The potential difference across both the cell will be equal
B.
C.
The potential difference across one of the cells will be greater than its emf.
D.
One of the cell, will supply energy to the other cell.

Solution:

(A) Potential difference across each cell = V_P – V_Q

(B) If i is clockwise then E₂ is source and for anti-clockwise current E₁ is source.

The correct answers are: The potential difference across both the cell will be equal, One of the cell, will supply energy to the other cell., The potential difference across one of the cells will be greater than its emf.,

*Multiple options can be correct

QUESTION: 4

In the figure shown: (All batteries are ideal)

A.
power supplied by 20V cell is –20W
B.
current through 25V cell is 20A
C.
power supplied by 20V cell is 20W
D.
current through 25V cell is 12.5A

Solution:

Power supplied by 20V cell = (–1)(20) = – 20W
as the cell is not supplying the power, it is eating the power (getting charged)

The correct answers are: current through 25V cell is 12.5A, power supplied by 20V cell is –20W

*Multiple options can be correct

QUESTION: 5

A conductor of truncated conical (frustum) is connected to a battery of emf ∈ as shown in the figure.

If at a section distance x from left end, electric field intensity, potential and rate of generation of heat per unit length are E, V and H respectively, then which of the following graphs(s) is/are correct?

Solution:

Let a be the radius of left end side cross-section, then radius of cross-section at distance x from left end is a + bx where b is a constant.

as i and σ are same for all cross-section

Rate of heat generation per unit length,

The correct answers are:

*Multiple options can be correct

QUESTION: 6

In the circuit shown in the figure

A.
potential difference across 4Ω resistance is equal to the potential difference across 6Ω resistance
B.
current flowing in the circuit is 5A
C.
current in wire AB is zero
D.
power supplied by the battery is 200 watt

Solution:

there is zero potential difference across 4Ω and 6Ω resistance.

power by battery P_b =

The correct answers are: power supplied by the battery is 200 watt, potential difference across 4Ω resistance is equal to the potential difference across 6Ω resistance

*Multiple options can be correct

QUESTION: 7

A variable current flows through a 1Ω resistor for 2 seconds. Time dependence of the current is shown in the graph.

A.
Total heat produced in the resistor is 50J.
B.
Maximum power during the flow of current is 100W.
C.
Average current through the resistor is 5A.
D.
Total charge flown through the resistor is 10C.

Solution:

Total charge = = Area under the curve = 10C

Maximum Power = I²R when I is maximum current = 100 × 1 = 100W

The correct answers are: Total charge flown through the resistor is 10C., Average current through the resistor is 5A., Maximum power during the flow of current is 100W.

*Multiple options can be correct

QUESTION: 8

Consider a resistor of uniform cross sectional area connected to a battery of zero internal resistance. If the length of the resistor is doubled by stretching it then :

A.
current will become four times.
B.
the thermal power produced by the resistor will become one fourth.
C.
the electric field in the wire will become half.
D.
the product of the current density and conductance will become half.

Solution:

As the length is doubled, the cross section are of the wire becomes half, thus the resistance of the wire becomes four times the previous value. Hence after the wire is elongated the current becomes one fourth. Electric field is potential difference per unit length and hence becomes half the initial value. The power delivered to resistance is and hence becomes one fourth.
The correct answers are: the electric field in the wire will become half., the thermal power produced by the resistor will become one fourth.

*Multiple options can be correct

QUESTION: 9

AB is part of a circuit as shown, that absorbs energy at a rate of 50W. E is an emf device that has no internal resistance.

A.
Rate of conversion from electrical to chemical energy is 48W in device E.
B.
Emf of the device is 48V.
C.
Potential difference across AB is 48V.
D.
Point B is connected to the positive terminal of E.

Solution:

Resistance absorbs energy at the rate of 2W.

Potential difference across AB $\Rightarrow$ V_AB·I = 50W

V_AB = 50V

Drop across resistor is 2V, therefore EMF of E is 48V.

As AB is absorbing energy at the rate of 50W, 48W is being absorbed by E. Thus E is on charging mode i.e. current is entering from +ve terminal of E.

The correct answers are: Emf of the device is 48V., Rate of conversion from electrical to chemical energy is 48W in device E.

*Multiple options can be correct

QUESTION: 10

The galvanometer shown in the figure has resistance 10Ω. It is shunted by a series combination of a resistance S = 1Ω and an ideal cell of emf 2V. A current 2A passes as shown.

A.
The potential difference across the resistance S is 1.5V
B.
The reading of the galvnometer is 1A
C.
The potential difference across the resistance S is 2V
D.
The reading of the galvnometer is zero

Solution:

Let the currents be as shown in the figure
KVL along ABCDA
⇒ –10i – 2 + (2 – i)1 = 0
∴ i = 0
Potential difference across S = (2 – i)1 = 2 × 1 = 2V