capacitor $x$ and $y$ are in parallel.So thin capacitance will be
$C_{n e t}=3+3=6 u F$
Capacitors $m$ and $N$ are in parallel Son this net capacitance will be
$C_{n e t}=1+2 u F$
So, the above diagram counted into
(Refer image $2$)
at $t=0 i=0$
Now, For $V_{A B}$
(Refer image $3$)
$V_{A}+\frac{150 u c}{6 u F}=V_{B}$
$V_{A}-V_{b}=-25 v=\left|V_{A}-V_{B}\right|=25 v$
$V_{A B}=25 V$
For $V_{B C}$
$V_{B}+\frac{150 u c}{2 u F}=V_{c}$
$V_{B}-V_{c}=-75 V$
$\Rightarrow\left|V_{B}-V_{c}\right|=75 V \rightarrow V_{B C}=75$
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