Discussion Forum

The supply voltage in a room is 120 V. The resistance of the lead wires is 6 Ω. A 60 W  bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

Answer:

Explanation:

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

$R= \frac{120\times120}{60}=240$  Ω

  R_eq = 240+6=246 Ω

$\Rightarrow$    $i_{1}=\frac{V}{R_{eq}}=\frac{120}{246}$

$v_{1}=\frac{240}{246}\times120=117.073 V$

  $\Rightarrow$   $i_{2}=\frac{120}{48+6}$

  $v_{2}=\frac{48}{54}\times 120=106.66V$

   v₁-v₂= 10.04 V