A bulb rated 36W and 12V is connected across a 20V cell. What resistan

1.	A bulb rated 36W and 12V is connected across a 20V cell. What resistance is required to glow it with full intensity.
Answer» 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. V = IR ⇒ V = I(R₁ + R₂) ⇒ 20 = 3(4 + R₂) ⇒ 4 + R₂ = 6.7 ⇒ R₂ = 6.7 - 4 ⇒ R₂ = 2.7 Ω 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.

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

Answer»

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.

V = IR

⇒ V = I(R₁ + R₂)

⇒ 20 = 3(4 + R₂)

⇒ 4 + R₂ = 6.7

⇒ R₂ = 6.7 - 4

⇒ R₂ = 2.7 Ω

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.