In the circuit below,C_1 = 20 , mu F,C_2 = 40 | vedclass

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(A) The capacitors are connected in series,so the charge $q$ on each capacitor is the same.Given the maximum voltage rating $V_{\max} = 50 \, V$ for e

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$95$

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(A) The capacitors are connected in series,so the charge $q$ on each capacitor is the same.Given the maximum voltage rating $V_{\max} = 50 \, V$ for each capacitor,the maximum charge each can hold is:$q_1 = C_1 V_{\max} = 20 \, \mu F 	imes 50 \, V = 1000 \, \mu C$$q_2 = C_2 V_{\max} = 40 \, \mu F 	imes 50 \, V = 2000 \, \mu C$$q_3 = C_3 V_{\max} = 50 \, \mu F 	imes 50 \, V = 2500 \, \mu C$Since they are in series,the circuit is limited by the capacitor with the smallest maximum charge capacity. Thus,$q_{\max} = 1000 \, \mu C$.Now,we calculate the voltage across each capacitor when $q = 1000 \, \mu C$:$V_1 = \frac{q}{C_1} = \frac{1000 \, \mu C}{20 \, \mu F} = 50 \, V$$V_2 = \frac{q}{C_2} = \frac{1000 \, \mu C}{40 \, \mu F} = 25 \, V$$V_3 = \frac{q}{C_3} = \frac{1000 \, \mu C}{50 \, \mu F} = 20 \, V$The total potential difference $V_{XY}$ is the sum of the individual potential differences:$V_{XY} = V_1 + V_2 + V_3 = 50 \, V + 25 \, V + 20 \, V = 95 \, V$.

In the circuit below,$C_1 = 20 \, \mu F$,$C_2 = 40 \, \mu F$,and $C_3 = 50 \, \mu F$. If no capacitor can sustain more than $50 \, V$,then the maximum potential difference between $X$ and $Y$ is .......... $V$.

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