Current Electricity(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

CURRENT ELECTRICITY MCQs

Electricity And Circuits

Total Questions : 60 | Page 4 of 6 pages

Question 31. A battery of emf E and internal resistance r is connected to a resistor of resistance

r_{1}

and Q joules of heat is produced in a certain time t. When the same battery is connected to another resistor of resistance

r_{2}

the same quantity of heat is produced in the same time t the value of r is

r21r2
r22r1
12(r1+r2)
√r1r2

Discuss Question

Answer: Option D. -> √r1r2
:
D
In the first case the current in the circuit is

l_{1} = \frac{E}{r_{1} + r}

Q_{1} = l_{2} r_{1} t = (\frac{E}{r_{1} + r})^{2} \times r_{1} \times t

In the second case

Q_{2} = (\frac{E}{r_{1} + r})^{2} \times r_{2} \times t

equating above equations we get

\frac{r_{1}}{(r_{1} + r)^{2}} = \frac{r_{2}}{(r_{2} + r)^{2}}

r_{1} (r_{2} + r)^{2} = r_{2} (r_{1} + r)^{2}

r_{1} (r_{2}^{2} + 2 r r_{2} + r^{2}) = r_{2} (r_{1}^{2} + 2 r r_{1} + r^{2})

r^{2} (r_{1} - r_{2}) = r_{1} r_{2} (r_{1} - r_{2})

r = \sqrt{r_{1} r_{2}}

Question 32. The current voltage (I – V) graphs for a given metallic wire at two different temperatures

T_{1}

and

T_{2}

are shown in figure. If follows from the graphs that

The Current Voltage (I – V) Graphs For A Given Metallic Wi...

T1 > T2
T1
T1 = T2
T1 is greater or less than T2 depending on whether the resistance R of the wire is greater or less than the ratio V/I

Discuss Question

Answer: Option B. -> T1
:
B

It is clear from figure that at a given voltage

V_{0}

the current

I_{1}

in the wire at temperature

T_{1}

is greater than the current

I_{2}

in the wire at temperature

T_{2}

, Therefore the resistance of the wire at temperature

T_{1}

is less than that of temperature

T_{2}

. This can happen if

T_{1}

is less than

T_{2}

because the resistance of wire increases with increase in temperature

Question 33. An electric kettle has two coils. When one coil is connected to the AC mains the water in the kettle boils in 10 minutes. When the other coil is used the same quantity of water takes 15 minutes to boil. How long will it take for the same quantity of water to boil if the two coils are connected in parallel?

6 min
12 min
18 min
24 min

Discuss Question

Answer: Option A. -> 6 min
:
A
Let H be the amount of the heat energy needed to boil the given quantity of water. If

R_{1}

and

R_{2}

are the resistances of coils and V is the applied voltage then

H = \frac{V^{2} t_{1}}{R_{1}} = \frac{V^{2} t_{2}}{R_{2}}

\frac{R_{2}}{R_{1}} = \frac{t_{2}}{t_{1}} = \frac{15}{10} = \frac{3}{2}

when the coils are connected in parallel the combined resistance is given by

\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{R_{1} + R_{2}}{R_{1} R_{2}}

R = \frac{R_{1} R_{2}}{R_{1} + R_{2}}

If the water takes t minutes to boil then

\frac{v^{2} t}{R} = \frac{V^{2} t_{1}}{R_{1}}

t = t_{1} \times \frac{R}{R_{1}} = t_{1} \times \frac{R_{1} R_{2}}{R_{1} + R_{2}} \times \frac{1}{R_{1}} = t_{1} \times \frac{1}{(\frac{R_{1}}{R_{2}} + 1)} = \frac{10 \times 1}{(\frac{2}{3} + 1)} = 6

min

Question 34. Cell supplies a current of 0.9 A through a 2 ohm resistor and current of 0.3 A through a 7 ohm resistor. What is the internal resistance of the cell?

0.5 ohm
1 ohm
1.2 ohm
2 ohm

Discuss Question

Answer: Option A. -> 0.5 ohm
:
A
If E the emf of the cell and r its internal resistance then

\frac{E}{2 + r} = 0.9

and

\frac{E}{7 + r} = 0.3

dividing the two equations we get

\frac{7 + r}{2 + r} = \frac{9}{3}

r=0.5ohm

Question 35. In the potentiometer circuit shown in figure the internal resistance of the 6 V battery is 1 ohm and the length of the wire AB is 100 cm. When AD = 60 cm the galvanometer shows no deflection. The emf of cell C is (the resistance of wire AB is 2 ohm)

0.7 V
0.8 V
0.9 V
1 V

Discuss Question

Answer: Option C. -> 0.9 V
:
C
Current in the circuit due to 6 V battery is

l = \frac{6}{1 + 5 + 2} = \frac{3}{4} A

Now emf of cell C =potential difference across AD

= \frac{3}{4} \times \frac{3 \times 60}{100} = 0.9 V

Question 36. An electric bell has a rating of 500 W 100 V. It is used in a circuit having a 200 V supply. What resistance must be connected in series with the bulb so that it delivers 500 W?

10 ohm
20 ohm
30 ohm
40 ohm

Discuss Question

Answer: Option B. -> 20 ohm
:
B
The current flowing in the bulb of 500Woperating at 100V is

I = \frac{500}{100} = 5 A

resistance of bulb

= \frac{100}{5} = 20

ohm say

R_{1}

to deliver 500W the current in the bulb must remain 5 A when it is operated with 200V supply the resistance R to be connected in series for this purpose is given by

\frac{200}{R_{1} + R_{2}} = 5

R_{1} + R_{2} = 40

R_{1} + 20 = 40

R=20ohm

Question 37. The current I in the circuit shown in figure is

145 A
115 A
110 A
15 A

Discuss Question

Answer: Option C. -> 110 A
:
C
The equivalent resistance is given by

\frac{1}{R} = \frac{1}{60} + \frac{1}{30} = \frac{3}{60}

R = 20ohm. Therefore

I = \frac{2}{20} = \frac{1}{10} A

Question 38. To replace a copper strip in a circuit, which among the following should we use to complete the circuit?

Wood
Rubber
Plastic
Silver

Discuss Question

Answer: Option D. -> Silver
:
D
Silver and copper are good conductors of electricity and allow electricity to pass through them easily. So, we can use silver strip instead of copper strip in a circuit. However, plastic and wood are poor conductors of electricity and do not allow electric current to pass through them easily. Hence, they cannot be used to replace copper.

Question 39. Which of the following appliances does not operate on an electric cell/battery?

Torch
TV remote
Walkie-talkie
Air conditioner

Discuss Question

Answer: Option D. -> Air conditioner
:
D
Usually, electric cells/batteries are capable of producing very less amount of electric current. Devices like torch, TV remote, and walkie-talkie require less amount of current to function and the use of electric cell is preferred for these applications.
On the other hand, an air conditioner requires larger current. Air conditioner uses mains power supply for its working.

Question 40. Electricity is very essential for our day to day activities.

True
False

Discuss Question

Answer: Option A. -> True
:
A
Electricityplays a huge part in our every day lives, whether it is at homes or our workplaces.
Our daily routines rely heavily on the use ofelectricity like using an iron, refrigerator, television, washing machine and so on.