Three capacitors are connected in the arms of triangle ABC shown in figure 5 V is applied\ between

Question

Three capacitors are connected in the arms of triangle ABC shown in figure 5 V is applied\ between A and B. The voltage between B and C is\

Professor PC Thomas Classes · Answer

The equivalent circuit diagram as show is the figure\ \ $\mathrm{C}_{	ext {eq }}=\frac{2 \mu \mathrm{~F} 	imes 3 \mu \mathrm{~F}}{2 \mu \mathrm{~F}+3 \mu \mathrm{~F}}+2 \mu \mathrm{~F}$\ The equivalent capacitance between A and B is\ $=\frac{6}{5} \mu \mathrm{~F}+2 \mu \mathrm{~F}=\frac{16}{5} \mu \mathrm{~F}$\ Total charge of the given circuit is\ $\mathrm{Q}=\frac{16}{5} \mu \mathrm{~F}+5 \mathrm{~V}=16 \mu \mathrm{C}$\ $\mathrm{Q}_{1}=(2 \mu \mathrm{~F}) 	imes 5 \mathrm{~V}=10 \mu \mathrm{C}$\ As $\mathrm{Q}=\mathrm{Q}_{1}+\mathrm{Q}_{2}$\ $	herefore \mathrm{Q}_{2}=\mathrm{Q}-\mathrm{Q}_{1}=16 \mu \mathrm{C}-10 \mu \mathrm{C}=6 \mu \mathrm{C}$\ $	herefore$ Voltage between B and C is\ $\mathrm{V}_{	ext {BC }}=\frac{\mathrm{Q}_{2}}{3 \mu \mathrm{~F}}=\frac{6 \mu \mathrm{C}}{3 \mu \mathrm{~F}}$\