Question

The power dissipated in the circuit shown in the figure is 30 watts. The value of $R$ is

(A) $15\Omega$
(B) $10\Omega$
(C) $30\Omega$
(D) $20\Omega$

Answer 1

The voltage across resistors in parallel is the same. The total power consumed is the sum of the power consumed by the individual resistor
Formula used: In this solution we will be using the following formulae;
$P = \dfrac{{{V^2}}}{R}$ where $P$ is the power consumed in a circuit, $V$is the voltage connected to the circuit, $R$ is the effective or equivalent resistance in the circuit.
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$ where ${R_{eq}}$ is the equivalent resistance of two resistors, and ${R_1}$ and ${R_2}$ are the individual resistances of the resistor.

Complete Step-by-Step Solution:
We are given the total power consumed in the circuit, and asked to find the resistance $R$.
We see that the resistance $R$ is connected in parallel to the resistance $5\Omega$. Hence, the voltage across both of them are equal. And the voltage is equal to 10 V.
The total power consumed in the circuit can be given as
$P = \dfrac{{{V^2}}}{R}$ where $P$ is the power consumed in a circuit, $V$is the voltage connected to the circuit, $R$ is the effective or equivalent resistance in the circuit.
We find the equivalent resistance in the circuit.
Equivalent resistance of a circuit can be given as
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$ where ${R_1}$ and ${R_2}$ are the individual resistances of the resistor.
Hence,
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{R} + \dfrac{1}{5}$
$\Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{{5 + R}}{{5R}}$
Hence, inverting it, we get
${R_{eq}} = \dfrac{{5R}}{{5 + R}}$
Hence, the power is given as
$P = \dfrac{{{V^2}}}{{\dfrac{{5R}}{{5 + R}}}}$
Hence,
$30 = \dfrac{{{{10}^2}}}{{\dfrac{{5R}}{{5 + R}}}} = \dfrac{{100(5 + R)}}{{5R}}$
Dividing both sides by 10, and cross multiplying, we have
$3R = 10 + 2R$
$R = 10\Omega$

Hence, the correct option is B.

Note: Alternatively, the total power consumed in the circuit is the sum of the power consumed in the individual resistance, hence, we can write that
$P = \dfrac{{{V^2}}}{{{R_5}}} + \dfrac{{{V^2}}}{R}$ where ${R_5}$ is the 5 ohms resistance.
If we factorize out ${V^2}$and put the value of the 5 ohms we have
$P = {V^2}\left( {\dfrac{1}{5} + \dfrac{1}{R}} \right)$
$P = {V^2}\left( {\dfrac{{5 + R}}{{5R}}} \right)$ which can be written as
$P = \dfrac{{{V^2}}}{{\dfrac{{5R}}{{5 + R}}}}$ which is identical to the formula in the solution.