A network of four capacitors, each of capacitance $24 μ F$ , is connected across a battery of 50 V, as shown in the figure. The net capacitance and the net energy stored in the capacitors is $\frac{x}{100}$ . Find the value of $x$ ?

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Solution

Here $C_{1}, C_{2}, C_{3}$ are in series and the equivalent resistance for these is $C_{123} = 24 / 3 = 8 μ F$
Now $C_{123}, C_{4}$ are in parallel. So the net capacitance $C_{n e t} = C_{123} + C_{4} = 8 + 24 = 32 μ F$
Thus, net energy stored in capacitors is $U_{n e t} = \frac{1}{2} C_{n e t} V^{2} = 0.5 \times 32 \times 10^{- 6} \times 50^{2} = 0.04 J$

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A network of four capacitors, each of capacitance 24μF, is connected across a battery of 50 V, as shown in the figure. The net capacitance and the net energy stored in the capacitors is x100. Find the value of x?

Here C1,C2,C3 are in series and the equivalent resistance for these is C123=24/3=8μFNow C123,C4 are in parallel. So the net capacitance Cnet=C123+C4=8+24=32μFThus, net energy stored in capacitors is Unet=12CnetV2=0.5×32×10−6×502=0.04J

A network of four capacitors, each of capacitance $24 μ F$ , is connected across a battery of 50 V, as shown in the figure. The net capacitance and the net energy stored in the capacitors is $\frac{x}{100}$ . Find the value of $x$ ?