Name: RC Circuit and Charging Capacitor
Uploaded: 2019-07-19
Duration: 13 min 36 s
Description: In this video, we will discuss the physics of charging the capacitor connected in a circuit and derive the expression and relation between the parameters inv...

Question

$1 kW, 220 V$ electric heater is to be used with $220 V D . C .$ supply.

(a) What is the current in the heater?

(b) What is its resistance?

(c) What is the power dissipated in the heater?

(d) How much heat in calories is produced per second?

(e) How many grams of water at $100 ° C$ will be converted per minute into steam at $100 ° C$ ?

Accepted Answer

Given voltage to the heater, $V = 220 V$ ,

Power rating of the heater, $220 V D . C .$ ,

Now,

(a) We know that power generated by a resistive load with certain current, $I$ and voltage supply, $V$ is given as, $P = V I$ ,

So, $P = V \times I$

$\Rightarrow 1000 = 220 \times I$

$\Rightarrow I = \frac{1000}{220} = 4.55 A .$

Hence, the current in the heater is $4.55 A$ .

(b) Power generated in the heater can be given as, $P = \frac{V^{2}}{R} (R$ is the resistance of the heater $)$

$\Rightarrow 1000 = \frac{220 \times 220}{R}$

$\Rightarrow R = \frac{48400}{1000} = 48.4 Ω .$

Hence, the resistance of the heater is $48.4 Ω$ .

(c) As given power rating of the heater is $1000 W$ so this is the power dissipated in the heater which can also be given as, $P = \frac{V^{2}}{R} = \frac{220 \times 220}{48.4} = 1000 W$ .

(d) Heat, $H$ , produced per second of the heater, i.e. $H = \frac{V^{2}}{R} \times t = \frac{220 \times 220}{48.4} \times 1 = 1000 J$ .

Since $1 J = \frac{1}{4.2} calories,$

So, $1000 J = \frac{1}{4.2} \times 1000 = 238.1 calories .$

So, heat in calories produced per sec is $238.1 cal s^{- 1}$ .

(e) Heat produced per second is $238.1 calories s^{- 1}$ (as obtained above) so the same heat will be converted to steam and if taken $per minute$ then, we get, $238.1 \times 60 = 14286 calories \min^{- 1} .$

Since, the water at $100 {}^{o}C$ is converted to steam at $100 {}^{o}C$ , and we know that during phase change the heat energy is released or absorbed is given as, $Q = m L$ ,

Here $Q = 14286 calories,$

Latent heat of water $L = 540 cal g^{- 1}$

$m$ is the mass of water.

So, $m = \frac{Q}{L}$

$\Rightarrow m = \frac{14286}{540} = 26.5 g .$

Hence, mass of water at $100 {}^{o}C$ converted per minute into steam at $100 {}^{o}C$ is $26.5 g .$

Important Questions on Current Electricity

Learn and Prepare for any exam you want !