Current Electricity 5 Question 9
9. The supply voltage in a room is $120 V$. The resistance of the lead wires is $6 \Omega$. 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?
(2013 Main)
(a) zero
(b) $2.9 V$
(c) $13.3 V$
(d) $10.4 V$
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Solution:
- As, $P=\frac{V^{2}}{R}$
where, $P=$ power dissipates in the circuit,
$V=$ applied voltage,
$R=$ net resistance of the circuit
$R=\frac{120 \times 120}{60}=240 \Omega \quad$ [resistance of bulb]
$$ \begin{array}{rlr} R _{eq} & =240+6=246 \Omega & \\ \Rightarrow \quad i _1 & =\frac{V}{R _{eq}}=\frac{120}{246} \quad \text { [before connecting heater] } \\ R & =\frac{V^{2}}{R}=\frac{120 \times 120}{240} & \\ \Rightarrow \quad R & =60 \Omega & \text { [resistance of heater] } \end{array} $$
So, from figure,
$$ \begin{gathered} V _1=\frac{240}{246} \times 120=117.073 V \quad[\because V=I R] \\ \Rightarrow i _2=\frac{V}{R _2}=\frac{120}{54} \Rightarrow V _2=\frac{48}{54} \times 120=106.66 V \\ V _1-V _2=10.41 V \end{gathered} $$
