Question

A 100 W, 250 V lamp is connected in parallel with a resistance R across a supply voltage of 250 V. The total power in the circuit is 1100 W. Find the value of R

Answer 1

Answer:

$62.5 \Omega$

Explanation:

First of all, we calculate the resistance of the lamp. The power of the lamp is given by

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

Where

P = 100 W is the power

V = 250 V is the voltage

R' is the resistance of the lamp

Solving for R',

$R'=\frac{V^2}{P}=\frac{250^2}{100}=625 \Omega$

Now this resistance is connected in parallel to the other resistance R, so their equivalent resistance is

$\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R'}$ (1)

Now the total power in the circuit is

P = 1100 W

And the voltage is

V = 250 V

So we can write

$P=\frac{V^2}{R_{eq}}$

to find the equivalent resistance:

$R_{eq}=\frac{V^2}{P}=\frac{250^2}{1100}=56.8 \Omega$

Therefore, rearranging (1), we find R:

$\frac{1}{R}=\frac{1}{R_{eq}}-\frac{1}{R'}=\frac{1}{56.8}-\frac{1}{625}=0.016 \Omega^{-1} \rightarrow R=62.5 \Omega$