Question

Q.20. An electric bulb is rated at 60W, 240V. 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 remains
unchanged).

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Answer 1

Given:-

• Power of bulb ,P = 60 W

• Potential Difference ,V = 240 V

To Find:-

• Resistance ,R
• Power ,P
• Current ,I

Solution:-

Firstly we have to calculate the Resistance

• P = V²/R

Substitute the value we get

→ 60 = 240²/R

→ 60 = 240×240/R

→ 60 ×R = 240 ×240

→ R = 240×240/60

→ R = 960 Ohm

Therefore, the resistance of the bulb is 960 ohm.

Now Calculating the Power of bulb when potential difference is drop to 192 V

• P = V²/R

Substitute the value we get

→ P = 192² /960

→ P = 192×192/960

→ P = 38.4 Watt.

Therefore, the power of the bulb is 38.4 watt.

Now , Calculating the Current drawn by the bulb

• P = VI

Substitute the value we get

→ 38.4 = 192×I

→ I = 38.4/192

→ I = 0.2 A

Therefore, the current drawn by the bulb is 0.2 Ampere.