WBCHSE Class 12 Physics Electric Energy And Power Short Question And Answers

WBCHSE Class 12 Physics Electric Energy

Electric Energy And Power Short Question And Answers

Question 1. A parallel combination of three resistors 3Ω, and 5Ω is connected across a battery. Find which resistor will consume more electrical energy per second.
Answer:

As the three resistances are connected in parallel, the voltage drop across each of them will be the same.

∴ ‍Power dissipated in 3 Ω resistance, \(P_1=\frac{V^2}{3}\)

Similarly, \(P_2=\frac{V^2}{4} \text { and } P_3=\frac{V^2}{5}\)

So, the resistance 3Ω will consume more energy in 1 s.

Question 2. The rate of heat developed in a resistor R connected to a supply of potential difference V is H. What will be the rate of heat developed if the potential difference is V/3 and the resistance doubles?
Answers:

In the first case,

⇒ \(H \propto \frac{V^2}{R} \text { or, } H=k \frac{V^2}{R}\) [where k = constant]

In the second case,

⇒ \(H^{\prime}=k \frac{\left(\frac{V}{3}\right)^2}{2 R}=\frac{1}{18} k \frac{V^2}{R}=\frac{1}{18} H\)

Electric Energy and Power Short Questions 

Electric Energy Short Q&A WBCHSE

Question 3. A light bulb is rated 100 W for a 220 V ac supply of 50 Hz.

1. The resistance of the bulb,
2. Therms current through the bulb.

Answer:

200 V = rms supply voltage = equivalent dc voltage

1. \(P=\frac{V^2}{R} \quad \text { or, } R=\frac{V^2}{P}=\frac{(220)^2}{100}=484 \Omega\)

2. \(I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{R}=\frac{220}{484} \approx 0.45 \mathrm{~A}\)

Class 12 Physics Electric Power Questions 

Question 4. Nichrome and copper wires of the same length and radius are connected in series. Current A is passed through them. Which wire gets heated up more?
Answer:

Heat dissipates in a wire, H = I²Rt

We know that,

⇒ \(R=\frac{\rho l}{A}\)

∴ \(H=I^2 \frac{\rho l}{A} t\)

If current I, length l, and area A remain the same, H depends on \(\rho\)

∴ \(H \propto \rho \text { and } \rho_{\text {nichrome }}>\rho_{\text {copper }}\)

Hence, the nichrome wire will heat up more.

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WBCHSE Physics Questions on Electric Power

Question 5. The potential difference applied across a given resistor is altered so that the heat produced per second increases by a factor of 9. By what factor does the applied potential difference change?
Answer:

We know that, \(H=\frac{V^2}{R} t\)

or, \(\frac{H}{t}=\frac{V^2}{R} \quad ∴ \frac{H}{t} \propto V^2\)

It is given that heat produced per second \(\frac{H}{t}\), increases by a factor of 9.

Hence, the applied potential difference V increased by a factor of 3.

Electric Energy and Power Class 12 Notes 

Question 6. The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r. Obtain the expression for

1. The current drawn from the cell
2. The power consumed in the network.

Answer:

Here r, 2r, 2r, and r are in parallel.

So, \(\frac{1}{R_{A B}}=\frac{1}{r}+\frac{1}{r}+\frac{1}{2 r}+\frac{1}{2 r}\)

or, \(\frac{1}{R_{A B}}=\frac{3}{r} \quad \text { or, } R_{A B}=\frac{r}{3}\)

The total resistance of the circuit,

⇒ \(R=r+R_{A B}=r+\frac{r}{3}=\frac{4 r}{3}\)

1. Current drawn from the cell,

⇒ \(I=\frac{E}{R}=\frac{E}{(4 r / 3)}=\frac{3 E}{4 r}\)

2. Power consumed in network, \(P=I^2 R_{A B}\)

∴ \(P=\left(\frac{3 E}{4 r}\right)^2 \cdot \frac{r}{3}=\frac{3 E^2}{16 r}\)

WBCHSE Physics Chapter 3 Solutions 

Short Answer Questions on Electric Energy WBCHSE

Question 7. Two electric bulbs P and Q have their resistances In the ratio of 1: 2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs.
Answer:

We have, \(\frac{R_P}{R_Q}=\frac{1}{2}\)

Power consumed by a bulb,

P = I²R

Here, the current I is the same for both bulbs.

So, \(\frac{P_p}{P_Q}=\frac{R_p}{R_Q}=\frac{1}{2}\)