Question

Q1. An electric bulb is rated at 60 W, 240 V. Calculate its resistance. If the voltage drops to 192 V, calculate the power consumed and the current drawn by the bulb. (Assume that the resistance of the bulb remain unchanged?​

Answer 1

Answer:

0.2 A

Explanation:

Power rating =60w

Voltage rating = 240v

Resistance =v^ 2/p =

240× 240/60

R= 260 ohm

voltage across the bulb = 192 volt

power consumed =

voltage ^2/R=

192×192/960=384 watt

current across the bulb = voltage /R

= 192/960 =0.2 A.

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Answer 2

Given:  

Power= 60W

Voltage=240V

To Calculate:

Power consumed and Current drawn by bulb

solution:

Power is given by

P=$\frac{V^{2} }{R}$

∴R= $\frac{V^{2} }{P}$=$\frac{240^{2} }{60}$=4×240= 960Ω

Hence, the resistance of the bulb is 960 ohms.

Now, the voltage drops to 192V.

The resistance remains the same. So, we get

P=$\frac{V^{2} }{R}$=$\frac{192^{2} }{960}$=38.4W

Hence, the power consumed by the bulb is 38.4W.

Now, power and current are related as

Power=Voltage× current

∴ I= $\frac{p}{v}$=$\frac{38.4}{192}$= 0.2 A