Question

In a circuit given below, the percentage change in the total heat dissipated per unit time on closing the switch S is:

Answer 1

Joule's law of heating

$P=I^{2} R =  \frac{V^2}{R}$

Resistors connected in series, $R_{s} = R_1 +R_2$

Resistors connected in parallel, $\frac{1}{R_p} = \frac{1}{R_1} +\frac {1}{R_2}$

When Switch S is open,

$R_{s1} = 2+4 =6\\R_{s2} = 3+9 = 12$

Equivalent resistance of the circuit is

$\frac{1}{R_{eq}} =\frac{1}{R_{s1}} + \frac{1}{R_{s2}} \\ = \frac{1}{6} + \frac{1}{12} =\frac{2+1}{12} =\frac{1}{4} \\ \therefore R_{eq} = 4 \ohm$

Power dissipated, $P_{c}= \frac{V^2}{R_{eq}} = \frac{20^2}{4}=\frac{400}{4}=100 </p><p>When Switch S is closed,   </p><p>[tex]R_{s1} = 2+4 =6\\R_{s2} = 3+0 = 3$

Equivalent resistance of the circuit is

$\frac{1}{R_{eq}} =\frac{1}{R_{s1}} + \frac{1}{R_{s2}} \\ = \frac{1}{6} + \frac{1}{3} =\frac{1+2}{6} =\frac{1}{2} \\ \therefore R_{eq} = 2 \ohm$

Power dissipated, [tex]P_{c}= \frac{V^2}{R_{eq}} = \frac{20^2}{2}=\frac{400}{2}=200

Change in heat dissipated = 200 -100 =100

% change = \frac { 100}{200} * 100 = 50%