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Find the amount of heat generated in the circuit shown in the figure after the switch is shifted from position 1 to position 2.

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When the switch is in position 1, the combination has C and $C_0$ in parallel and C in series for which the equivalent capacitance is
$C_{eq}= \large\frac{C(C+C_0)}{2C+C_0}$
The total charge on the combination is
$Q= EC_{eq} = \large\frac{EC(C+C_0)}{2C+C_0}$
The total charge on the three capacitors can be obtained as
Now, heat produced = loss in stored electrical energy + extra energy drawn from the battery. Since the equivalent capacitance Ceq remains uncharged in both the positions of the key, the loss in stored energy is zero.
Hence, Heat produced = energy drawn from the battery
$\quad= E. \Delta q= E(q'_1-q_1)$
$\quad= E(q_3 -q'_3)$
$\quad=\large\frac {E^2 CC_0}{2C+C_0}$
answered Jun 19, 2014 by meena.p
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