Question

An electric bulb is rated at 60w , 240v. calculate its resistance .if the voltage drops to 192 v calculate the power consume and the current drawn by the bulb. (assume that the resistance of the bulb remain uncharged).

Answer 1

Hey friend, Harish here.

Here is your answer:

Given that:

i) Rated Power of bulb (p) = 60 w.

ii) Rated Voltage (v) = 240 v.

iii) Voltage drop across battery (v') = 192 v.

To find,

The resistance , power consumed and the current drawn by the bulb.

Solution:

We know that,

$Resistance (R) = \frac{V^{2}}{P}$

⇒ $R = \frac{(240)^{2}}{60}$

⇒ $R = 960\ \Omega$

Voltage drop across the battery  = 192 V.

Let power consumed be ( p' ).

Then, 

 $p' = \frac{(v')^{2}}{R}$

⇒ $p' = \frac{(192)^{2}}{960}$

⇒ $p' = 38.4\ W$

Then, Current across the battery is (I):

⇒ $I = \frac{V'}{R} = \frac{192}{960} = 0.2\ A$

$\bold{Therefore \ \boxed{R = 960 \Omega\ ;\ P' =38.4\ W\ ;\ I = 0.2 A } }$
____________________________________________________

Hope my answer is helpful to you.

Answer 2

Given
Power of electric bulb (P) = 60 W
The potential difference of the bulb (V) = 240 V
The resistance of the bulb (R) =?
We know
P = V²/R
60 = (240)²/R
R = 57600/60
R = 960 ohm

The resistance of the bulb = 960 ohm
Now, the voltage of the bulb falls to 192 V
Resistance remains same
We know V = IR
192 = I * 960
I = 192/960
I = 0.2 A
The current of the bulb = 0.2 A
Now,
The power of the bulb = Current * Voltage of the bulb
P = 0.2 * 192
P = 38.4 W

The power of the bulb = 38.4 W

Hope this helps you.