Answer

Verified

415.5k+ views

**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.

Recently Updated Pages

In the given figure ABDE Show that angle x + angle class 9 maths CBSE

In the given figure ABC is a triangle The bisector class 9 maths CBSE

In the given figure ABC is a triangle in which AB -class-9-maths-CBSE

In the given figure ABCD is rhombus and ALMC is square class 9 maths CBSE

In the given figure ABCD is a trapezium in which ABparallel class 9 maths CBSE

In the given figure ABCD is a square of side 7 cm and class 9 maths CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Name 10 Living and Non living things class 9 biology CBSE

The Buddhist universities of Nalanda and Vikramshila class 7 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE