Current Electricity - Result Question 67

72. The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is: [2012M]

(a) $20 \Omega$

(b) $15 \Omega$

(c) $10 \Omega$

(d) $30 \Omega$

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Answer:

Correct Answer: 72. (c)

Solution:

  1. (c) The power dissipated in the circuit.

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

$V=10$ volt

$\frac{1}{R _{eq}}=\frac{1}{R}+\frac{1}{5}=\frac{5+R}{5 R}$

$R _{eq}=(\frac{5 R}{5+R})$

$P=30 W$

Substituting the values in equation (i)

$30=\frac{(10)^{2}}{(\frac{5 R}{5+R})}$

$\frac{15 R}{5+R}=10$

$15 R=50+10 R$

$5 R=50$

$R=10 \Omega$

For a given voltage $V$, if resistance is changed from $R$ to $(\frac{R}{n})$. Power consumed changes from $P$ to $n P$.

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

when $R^{\prime}=\frac{R}{n}$

then $P^{\prime}=\frac{V^{2}}{R / n}=\frac{n V^{2}}{R}=n P$