Question

In the circuit shown, the resistance are given in ohms and the battery is assumed ideal with emf equal to 3.0 volts. The resistor that dissipates the most power is-

• A. $R_1$
• B. $R_2$
• C. $R_3$
• D. $R_4$

Answer 1

The correct option is A $R_1$

Given circuit can be simplified as,

$V_1=\frac{50}{50+\frac{100}{7}}\times 3\Rightarrow \frac{150}{450}\times 7=\frac{7}{3}V$
$V_2=3-\frac{7}{3}=\frac{2}{3}V$
Power dissipated through resistance $R_1$ is
power $P_1=\frac{V_1^2}{R_1}=\frac{49}{9\times 50}=\frac{49}{450}V$

$R_2,R_3$ and $R_4$ have same potential and $R_4$ is least among them
$\therefore$ Power in $R_4>$ Power in $R_2,R_3$
Power dissipated through resistance $R_4$ is
$\therefore$ Power $P_4=\frac{V_2^2}{R_4}$
Power $P_4=\frac{4}{9\times 30}=\frac{4}{270}V$
$\therefore$ Power dissipated in $R_1$ is max.