Current Electricity Test

Result

Current Electricity Test - 133

Score
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Weekly Quiz Competition

Question 1
4 / -1

A
2 Ω and 0.5 Ω

B
1 Ω and 0.25 Ω

C
8 Ω and 2 Ω

D
4 Ω and 1 Ω

Solution
Question 2
4 / -1

Three identical resistors connected in series with a battery, together dissipate 10W of power. What will be the power dissipated, if the same resistors are connected in parallel across the same battery?

A
60W

B
30 W

C
90 W

D
120 W

Solution
Question 3
4 / -1

A potentiometer wire of length 1 m and resistance 10Ω is connected in series with a cell of e.m.f 2V with internal resistance 1Ω and a resistance box of a resistance R if potential difference between ends of the wire is 1V the value of R is.

A
4.5 Ω

B
9 Ω

C
15 Ω

D
20 Ω

Solution
Question 4
4 / -1

For a cell of e.m.f 2V, a balance is obtained for 50 cm of the potentiometer wire If the cell is shunted by a 2Ω resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell is.

A
1 Ω

B
0.5 Ω

C
1.2 Ω

D
2.5 Ω

Solution
Question 5
4 / -1

n idential cells each of e.m.f E and internal resistance r are connected in series An external resistance R is connected in series to this combination the current through R is.

A

B

C

D

Solution
Question 6
4 / -1

4 cell each of emf 2v and internal resistance of 1Ω are connected in parallel to a load resistor of 2Ω Then the current through the load resistor is....

A
2A

B
1.5 A

C
1A

D
0.888A

Solution
Question 7
4 / -1

A Potentiometer wire, 10m long, has a resistance of 40Ω It is connected in series with a resislance box and a 2V storage cell If the potential gradient along the wire is 0.1mv/cm, the resistance unplugged in the box is.

A
260 Ω

B
760 Ω

C
960 Ω

D
1060 Ω

Solution
Question 8
4 / -1

A

B

C

D

Solution
Question 9
4 / -1

Potentiometer wire of length 1m is connected in series with 490Ω resistance and 2V battery If 0.2 mv/ cm is the potential gradient, then resistance of the potentiometer wire is.

A
4.9 Ω

B
7.9 Ω

C
5.9 Ω

D
6.9 Ω

Solution
Question 10
4 / -1

A

B

C

D

Solution