Find the breakdown voltage of the given combi | vedclass

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(B) The circuit consists of two parallel branches. Branch $1$ has two capacitors in series: $C_1 = 6 \mu F$ (breakdown $V_1 = 1 	ext{ kV}$) and $C_2

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$4$

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(B) The circuit consists of two parallel branches. Branch $1$ has two capacitors in series: $C_1 = 6 \mu F$ (breakdown $V_1 = 1 	ext{ kV}$) and $C_2 = 2 \mu F$ (breakdown $V_2 = 4 	ext{ kV}$). The equivalent capacitance is $C_{eq1} = \frac{6 	imes 2}{6 + 2} = 1.5 \mu F$. The maximum voltage $V_{max1}$ for this branch is determined by the capacitor that reaches its breakdown limit first. For $C_1$,the charge is $q_1 = C_1 V_1 = 6 	imes 1 = 6 \mu C$. For $C_2$,the charge is $q_2 = C_2 V_2 = 2 	imes 4 = 8 \mu C$. Since they are in series,the charge $q$ must be the same. The limit is $q_{max1} = 6 \mu C$. Thus,$V_{max1} = \frac{q_{max1}}{C_{eq1}} = \frac{6}{1.5} = 4 	ext{ kV}$. Branch $2$ has two capacitors in series: $C_3 = 3 \mu F$ (breakdown $V_3 = 4 	ext{ kV}$) and $C_4 = 1 \mu F$ (breakdown $V_4 = 5 	ext{ kV}$). The equivalent capacitance is $C_{eq2} = \frac{3 	imes 1}{3 + 1} = 0.75 \mu F$. For $C_3$,$q_3 = 3 	imes 4 = 12 \mu C$. For $C_4$,$q_4 = 1 	imes 5 = 5 \mu C$. The limit is $q_{max2} = 5 \mu C$. Thus,$V_{max2} = \frac{q_{max2}}{C_{eq2}} = \frac{5}{0.75} = 6.67 	ext{ kV}$. The breakdown voltage of the entire parallel combination is the minimum of the two branch voltages: $V_{total} = \min(4, 6.67) = 4 	ext{ kV}$.

Find the breakdown voltage of the given combination in $KV$.

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