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A bulb rated 36W and 12V is connected across a 20V cell. What resistance is required to glow it with full intensity.

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+1 vote
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Let the resistance of the bulb be R

R = V2/ P

R = 122 / 36 = 4Ω 

Now, new V1 = 20 

R1 = V12 / P

R1 = 400/ 36 = 11Ω

Required resistance = 11Ω - 4Ω = 7Ω

Thus, 7Ω is required to glow the bulb with full intensity. 

–1 vote
by (19.8k points)

V = IR

⇒ V = I(R₁ + R₂)

⇒ 20 = 3(4 + R₂)

⇒ 4 + R₂ = 6.7

⇒ R₂ = 6.7 - 4

⇒ R₂ = 2.7 Ω

–1 vote
by (42.2k points)

Given :

Voltage of Bulb, V' = 12V

Power of Bulb, P = 36W

Potential of the Cell, V = 20V

To Find :

Resistance Required to Glow the Bulb with Full Intensity.

Resistance of the Given Bulb :

⇒ R₁  = (V')²/P

⇒ R₁  = (12)²/36

⇒ R₁  = 144/36

⇒ R₁  = 4Ω

Current passes through the Bulb :

⇒ I = P/V'

⇒ I = 36/12

⇒ I = 3A

It is a Series Connection. So that, The Current passes through the Whole Circuit is 3A .

Equivalent Resistance R = (R₁ + R₂)

Now, Substituting the Values in Ohm's Law :

⇒ V = IR

⇒ V = I(R₁  + R₂)

⇒ 20 = 3(4 + R₂)

⇒ 3R₂ = 20 - 12

⇒ R₂ = 8/3

⇒ R₂ = 2.67Ω

The Resistance of 2.7Ω is Required to Glow it with Full Intensity.

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