STD 12 - 2. Electric potential and capacitance — NEET STD 11 Science — Question

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 $V_{B C}$

$V_{B}+\frac{150 u c}{2 u F}=V_{c}$

$V_{B}-V_{c}=-75 V$

$\Rightarrow\left|V_{B}-V_{c}\right|=75 V \rightarrow V_{B C}=75$