SATHEE: Current Electricity - Result Question 67

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) $20 Ω$

(b) $15 Ω$

(d) $30 Ω$

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Answer:

Correct Answer: 72. (c)

Solution:

$P = \frac{V^{2}}{R_{e q}}$

$V = 10$ volt

$\frac{1}{R_{e q}} = \frac{1}{R} + \frac{1}{5} = \frac{5 + R}{5 R}$

$R_{e q} = (\frac{5 R}{5 + R})$

$P = 30 W$

Substituting the values in equation (i)

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

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

$15 R = 50 + 10 R$

$5 R = 50$

$R = 10 Ω$

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 $R^{'} = \frac{R}{n}$

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

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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]

Answer:

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