Question

A problem related to electricity and about resistance to be connected to a bulb. On using the current in the circuit to find the resistance, I get 20Ω which is the right answer. But on using the formula P=V²/(R+r), I get r=60Ω. What's wrong in using the second method?

Answer 1

Given:

In Case 1

Power(W)=500W

Potential Difference(V) =100V

In Case 2

Power(W)=500W

Potential Difference(V) = 200v

To Find:

Extra resistance that must be put in series with the bulb.

Solution:

In Case 1

Resistance of the bulb(R) = $\sf{\dfrac{V^2}{P}}$

$\sf{R=}{\dfrac{100^2}{500}}$

$\sf{R=20}$Ω

In Case 2

Total current in the circuit:

$\sf{V=200V$

$\sf{P=500W}$

$\sf{R=20}$Ω

$\sf{I=}{\sqrt\dfrac{P}{R}}$

$\sf{I=}{\sqrt\dfrac{500}{20}}$

$\sf{I=5A}$

Total resistance in the circuit:

From Ohm's Law:

$\sf{V=IR}$

$\sf{\therefore R=}{\dfrac{V}{I}}$

$\sf{R=}{\dfrac{200}{5}}$

$\sf{R=40}$Ω

Therefore, the extra resistance that must be put in series with the bulb is 20Ω.