Question

Two 32W-100V bulbs are connected in (i) in series (ii) in parallel to 100V supply. The power consumed by each in two cases is
A. $16W,64W$
B. $8W,32W$
C. $32W,32W$
D. $16W,32W$

Answer 1

Hint: The resistance of the bulb is given as$R=\dfrac{{{V}^{2}}}{P}$, where $V$is the power rating of the bulb and $P$ is the power rating of the bulb. In series combination the electric current in both the bulbs is the same and there is distribution of the electric current in the parallel connection in the bulb.

Complete step by step solution:
Rated power of each bulb is $P=32W$
Rated voltage of each bulb is $V=100V$
Hence, the resistance of each bulb can be calculated as,
$\begin{align}
  & R=\dfrac{{{\left( 100 \right)}^{2}}}{32}\Omega \\
 & =312.5\Omega \\
\end{align}$

Using Ohm’s law,
$\begin{align}
  & V=iR \\
 & i=\dfrac{V}{R} \\
\end{align}$
Where,
$i$ = electric current in the circuit
$V$ = Voltage applied in the circuit
$R$ = Net resistance in the circuit

Net resistance in circuit is given as,
$\begin{align}
  & {{R}_{series}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}\ldots \ldots +{{R}_{n}} \\
 & \dfrac{1}{{{R}_{Parallel}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}\ldots \ldots +\dfrac{1}{{{R}_{n}}} \\
\end{align}$

(I) In series combination,
${{R}_{series}}=\left( 312.5+312.5 \right)\Omega =625\Omega $
$i=\dfrac{100}{625}A=0.16A$

Then power consumed in each bulb is given as $P={{i}^{2}}R$
$P={{\left( 0.16 \right)}^{2}} \centerdot 312.5W=8W$

(II) In parallel combination,
$\begin{align}
  & \dfrac{1}{{{R}_{parallel}}}=\left( \dfrac{1}{312.5}+\dfrac{1}{312.5} \right)\dfrac{1}{\Omega }=\dfrac{2}{312.5\Omega } \\
 & {{R}_{parallel}}=\dfrac{312.5\Omega }{2}=156.25\Omega \\
\end{align}$
$i=\left( \dfrac{100}{156.25} \right) A=0.64A$

As the resistance of each bulb is the same, therefore electric current is equally distributed in each branch.
Hence, electric current in each resistor is ${{i}_{1}}={{i}_{2}}=\dfrac{i}{2}=\dfrac{0.64A}{2}=0.32A$
Power consumed in each bulb,
$\begin{align}
  & P=i_{1}^{2}R=i_{2}^{2}R={{\left( 032 \right)}^{2}}\centerdot 312.5W \\
 & P=32W \\
\end{align}$

Hence, the correct answer is option B.

Note: Electric current in series combination is same in all the resistors in series.
Electric current in parallel combination gets distributed among the parallel branches.