Question

In the circuit shown below the resistances 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

Answer: (A)

$R_1$

The effective resistance between the terminals of the battery is given by

$\text{series(40,parallel(40,series(parallel(60,30),20)))}$

$\text{parallel(60,30)}=\frac{60\times 30}{60+30}=20\Omega$

$\Rightarrow \text{series(parallel(60,30),20)=series(20,20) =}40\Omega$

$\Rightarrow \text{parallel(40,series(parallel(60,30),20))=parallel(40,40)=}20\Omega$

Thus, Equivalent resistance of the resistor network $=\text{series(40,20)}=60\Omega$

Total current through the cell = $\frac{V}{R_{eq}}=\frac{3}{60}=0.05A$

Current through $R_1=0.05A$

Resistance of $R_2=$ Resistance of $R_3{R}_4{R}_5$ network.

$\Rightarrow$ Current through $R_2=\frac{0.05}{2}=0.025A$

and Current through $R_3{R}_4$ network = Current through $R_5=0.025A$

In the $R_3{R}_4$ network, current splits in the inverse ratio of resistance.

$\Rightarrow$ Current in $R_3=\frac{0.025}{3}A$ and current in $R_4=\frac{0.05}{3}A$

Power dissipated in a resistor $\left(P\right)=I^{2}R$

$P_1={0.05}^2\times 40=0.1W$

$P_2={0.025}^2\times 40=0.025W$

$P_3=(\frac{0.025}{3}{)}^2\times 60=0.0041667W$

$P_4=(\frac{0.05}{3}{)}^2\times 30=0.00833W$

$P_5={0.025}^2\times 20=0.0125W$

Thus maximum power is dissipated in the resistor $R_1$