Current Electricity Test

Result

Current Electricity Test - 1

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

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
0.50 V

B
1.25 V

C
0.75 V

D
2.0 V

Solution

Explanation:

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

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

A
1.25 V

B
0.50 V

C
0.75 V

D
2.0 V

Solution

Explanation:

The cells are in opposition. Eeq=E₁−E₂=1.25−0.75=0.5V
Question 3
1 / -0

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

A
a slide wire bridge

B
an ammeter

C
a potentiometer

D
a voltmeter

Solution

Explanation:

Both potentiometer and voltmeter are devices to measure potential difference. E M F 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 4
1 / -0

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

A
it has a wire of low resistance.

B
it has a wire of high resistance

C
It draws a heavy current from external circuit.

D
It does not draw current from external circuit.

Solution

Explanation:

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 5
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
decreasing the current due to auxiliary battery

B
putting a suitable resistance in series with the cell.

C
Putting a shunt resistance in parallel with the cell.

D
increasing the current due to auxiliary battery

Solution

Explanation:

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 6
1 / -0

The sensitivity of the potentiometer can be increased by:

A
increasing the length of potentiometer wire.

B
decreasing the length of potentiometer wire.

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

D
increasing the potential gradient.

Solution

Explanation:

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 7
1 / -0

Kirchhoff’s second law is a consequence of

A
Law of conservation of charge

B
Law of conservation of angular momentum

C
Law of conservation of momentum

D
Law of conservation of energy

Solution

Explanation:

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.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Current Electricity Test - 1

Result

Current Electricity Test - 1

Score

-

Rank

-

SHARING IS CARING

Question 1

0.50 V

1.25 V

0.75 V

2.0 V

Solution

Question 2

1.25 V

0.50 V

0.75 V

2.0 V

Solution

Question 3

a slide wire bridge

an ammeter

a potentiometer

a voltmeter

Solution

Question 4

it has a wire of low resistance.

it has a wire of high resistance

It draws a heavy current from external circuit.

It does not draw current from external circuit.

Solution

Question 5

decreasing the current due to auxiliary battery

putting a suitable resistance in series with the cell.

Putting a shunt resistance in parallel with the cell.

increasing the current due to auxiliary battery

Solution

Question 6

increasing the length of potentiometer wire.

decreasing the length of potentiometer wire.

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

increasing the potential gradient.

Solution

Question 7

Law of conservation of charge

Law of conservation of angular momentum

Law of conservation of momentum

Law of conservation of energy

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.