Current Electricity Test

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Current Electricity Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

Two cells of 1.25 V and 0.75 V are connected in series with anode of one connected to cathode of the other. The effective voltage will be

A
2.0 V

B
0.75 V

C
1.25 V

D
0.50 V

Solution

The cells support each other.
E_eq= E₁+ E₂= 1.25 + 0.75 = 2.0V
Question 2
1 / -0

The instrument for the accurate measurement of the e.m.f of a cell is

A
a slide wire bridge

B
a voltmeter

C
an ammeter

D
a potentiometer

Solution

Both potentiometer and voltmeter are devices to measure potential difference. Emf is the terminal p.d between the electrodes of a cell in open circuit, i.e., when no current is drawn from it. Potentiometer measures the potential difference using null deflection method, where no current is drawn from the cell; whereas voltmeter needs a small current to show deflection. So, accurate measurement of p.d is done using a potentiometer.
Question 3
1 / -0

Potentiometer measures the potential difference more accurately than a voltmeter, because

A
It does not draw current from external circuit.

B
It has a wire of low resistance.

C
It has a wire of high resistance.

D
It draws a heavy current from external circuit.

Solution

Potentiometer measures the potential difference using null deflection method, where no current is drawn from the cell; whereas voltmeter needs a small current to show deflection. So, accurate measurement of p.d is done using a potentiometer.
Question 4
1 / -0

In a potentiometer experiment, for measuring internal resistance of a cell, the balance point has been obtained on the fourth wire. The balance point can be shifted to fifth wire by

A
increasing the current due to auxiliary battery

B
putting a shunt resistance in parallel with the cell

C
decreasing the current due to auxiliary battery

D
putting a suitable resistance in series with the cell

Solution

If the current due to the auxillary battery is decreased, the potential gradient will be decreases, so the balancing length increases.
Thus, null point will move to fifth wire.
Question 5
1 / -0

The sensitivity of the potentiometer can be increased by:

A
increasing the length of potentiometer wire.

B
increasing the potential gradient.

C
increasing the e.m.f. of primary cell.

D
decreasing the length of potentiometer wire.

Solution

A potentiometer is considered to be sensitive if the potential gradient dV/dl is low. Such a potentiometer can measure very small changes in potential difference. Increasing the length of the potentiometer wire decreases the potential gradient. Its sensitivity increases. Increasing potential gradient decreases the sensitivity, increasing the emf of the primary cell and by decreasing the length, potential gradient increases.
Question 6
1 / -0

Kirchhoff’s second law is a consequence of law of conservation of

A
charge

B
angular momentum

C
momentum

D
energy

Solution

Work done in moving a charge in a closed circuit is zero, therefore the algebraic sum of potential differences in a closed circuit is zero.

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Current Electricity Test - 8

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Current Electricity Test - 8

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Question 1

2.0 V

0.75 V

1.25 V

0.50 V

Solution

Question 2

a slide wire bridge

a voltmeter

an ammeter

a potentiometer

Solution

Question 3

It does not draw current from external circuit.

It has a wire of low resistance.

It has a wire of high resistance.

It draws a heavy current from external circuit.

Solution

Question 4

increasing the current due to auxiliary battery

putting a shunt resistance in parallel with the cell

decreasing the current due to auxiliary battery

putting a suitable resistance in series with the cell

Solution

Question 5

increasing the length of potentiometer wire.

increasing the potential gradient.

increasing the e.m.f. of primary cell.

decreasing the length of potentiometer wire.

Solution

Question 6

charge

angular momentum

momentum

energy

Solution

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