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  • In the circuit given in Fig. switch S is at position 1 for long time. Find the total heat generated in resistor of resistance <img alt="\mathrm{\left(2 r_0\right)}" src="https://entrancecorner.onco

In the circuit given in Fig. switch S is at position 1 for long time. Find the total heat generated in resistor of resistance \mathrm{\left(2 r_0\right)}, when the switch S is shifted from position 1 to position 2
 

Option: 1

\mathrm{\frac{C_0 E_0^2}{2}}


Option: 2

\mathrm{C_0 E_0^2}


Option: 3

\mathrm{\frac{C_0 E_0^2}{3}}


Option: 4

None


Answers (1)

best_answer

Step I: when the switch is at position 1: Since circuit is in the steady state, the current through circuit is zero


According to the loop rule, \mathrm{E_0-\frac{q_0}{C_0}=0 \Rightarrow q_0=C_0 E_0}
Step II: When the switch is at position 2: In this case, the total energy stored on the capacitor appears as heat energy in the resistor

\begin{aligned} &\mathrm{ \therefore \Delta H=I^2 R T, \therefore \Delta H \propto R }\\ &\mathrm{ \therefore \frac{\Delta H_1}{\Delta H_2}=\frac{R_1}{R_2}=\frac{r_0}{2 r_0}=\frac{1}{2} \Rightarrow \Delta H_2=2 \Delta H_1}\\ &\mathrm{But ~\Delta H=\Delta H_1+\Delta H_2=\frac{\Delta H_1}{2}+\Delta H_2=\frac{3}{2} \Delta H_2}\\ &\mathrm{\therefore \Delta H_2=\frac{2}{3} \Delta H=\frac{2}{3} \times \frac{1}{2} C_0 E_0^2=\frac{1}{3} C_0 E_0^2}\\ \end{aligned}

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manish

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