Question

For ensuring dissipation of same energy in all three resistors $(R_1,R_2,R_3)$ connected as shown in figure, their values must be related as

• A. $R_1=R_2=R_3$
• B. $R_2=R_3$ and $R_1=4R_2$
• C. $R_2=R_3$ and $R_1=1/4 R_2$
• D. $R_1=R_2+R_3$

Answer 1

Answer: (C)

$R_2=R_3$ and $R_1=1/4 R_2$

Answer 2

When resistors are connected in parallel potential difference across them is same. In the given circuit the resistor's $R_{2}$ and $R_{3}$ are connected in parallel hence potential difference $(V)$ across them is same. In order that they undergo same energy loss, $H=\frac{V^{2}}{R} t$, $R_{2}$ must be equal to $R_{3}$. i.e., $R_{2}=R_{3}$ Now resistor $R_{1}$ is in series with $R_{2}$, hence energy through them is $H=i^{2} R_{1} t=i_{1}^{2} R_{2} t$ where $i_{1}$ is current across $R_{2}$. Since $R_{2}=R_{3}$, therefore current through them is $\frac{i}{2}=i_{1}$ $\therefore i^{2} R_{1} t=\frac{i^{2}}{4} R_{2} t$ $\Rightarrow R_{1}=\frac{R_{2}}{4}$