Question 675: Current Electricity

Question

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

Accepted Answer

The power dissipated in the circuit
$$P = {{{V^2}} \over {{R_{eq}}}}$$    ...(i)
v = 10 volt
$${1 \over {{R_{eq}}}} = {1 \over R} + {1 \over 5} = {{5 + R} \over {5R}}$$
$${R_{eq}} = \left( {{{5R} \over {5 + R}}} \right)$$
P = 30 W
Substituting the values in equation (i)
$$30 = {{{{\left( {10} \right)}^2}} \over {\left( {{{5R} \over {5 + R}}} \right)}}$$
$${{15R} \over {5 + R}} = 10$$
15R = 50 + 10R
5R = 50
R = 10 $$\Omega $$

Answer

20 $$\Omega $$

Answer

15 $$\Omega $$

Answer

10 $$\Omega $$

Answer

30 $$\Omega $$

Current Electricity

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

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