Question

A 100W, 250V lamp is connected in parallel with an unknown resistance ‘R,’ across a 250 V supply. The total power dissipated in the circuit is 1100V. Find the value of unknown resistance. Assume the resistance of the lamp remains unaltered. 

Answer 1

Answer

Resistance of given lamp

$R=\frac{V^2}{P}=\frac{250\times250}{100}=625\Omega$

Now let unknown resistance is R

So equivalent of these resistance

$R_eq=\frac{625R}{625+R}$

Now total power

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

$1100=\frac{250\times250(625+R)}{625R}$

Now solute

$11R=625+R$

$R=56.8\Omega$