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(C) Let the energy dissipated in each resistor be $H$. Since the resistors ${R_2}$ and ${R_3}$ are connected in parallel,the potential difference acro

Vedclass · Accepted Answer

${R_2} = {R_3}$ and ${R_1} = \frac{1}{4}{R_2}$

Vedclass · Answer

(C) Let the energy dissipated in each resistor be $H$. Since the resistors ${R_2}$ and ${R_3}$ are connected in parallel,the potential difference across them is the same. Therefore,the energy dissipated in them is given by $H = \frac{V^2}{R}t$. For the energy to be the same,we must have ${R_2} = {R_3}$.Let the total current flowing through the circuit be $i$. The current ${R_1}$ is $i$. The current splits into ${i_1}$ and ${i_2}$ at the junction. Since ${R_2} = {R_3}$,the current splits equally,so ${i_1} = {i_2} = \frac{i}{2}$.The energy dissipated in ${R_1}$ is $H = i^2 {R_1} t$.The energy dissipated in ${R_2}$ is $H = i_1^2 {R_2} t = (\frac{i}{2})^2 {R_2} t = \frac{i^2}{4} {R_2} t$.Since the energy dissipated in all resistors is the same,we equate the energy in ${R_1}$ and ${R_2}$:$i^2 {R_1} t = \frac{i^2}{4} {R_2} t$${R_1} = \frac{{R_2}}{4}$.

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

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