A 50 W bulb connected in series with a heater coil is put to an AC mains. Now the bulb is replaced by a 100 W bulb. The heater output will

Question

A 50 W bulb connected in series with a heater coil is put to an AC mains. Now the bulb is replaced by a 100 W bulb. The heater output will (A) Double (B) Halve (C) Increase (D) Decrease

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/02d0981af60c1f9d51457b450eb701c6-.png" /> P1=((E/R+R'))2 R' P2=(E R'/((R)2+R')2) P2 > P1

Q. A $50 W$ bulb connected in series with a heater coil is put to an $A C$ mains. Now the bulb is replaced by a $100 W$ bulb. The heater output will

Double

Halve

Increase

Decrease

$P_{1} = (\frac{E}{R + R ^{'}})^{2} R^{'}$
$P_{2} = \frac{E R ^{'}}{( \frac{R}{2} + R ^{'} ) ^{2}}$
$P_{2} > P_{1}$

Q. A 50W bulb connected in series with a heater coil is put to an AC mains. Now the bulb is replaced by a 100W bulb. The heater output will