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Q.
In the combination of capacitors shown in figure the potential difference across the plates of the capacitor $A$ will be
Electrostatic Potential and Capacitance
Solution:
In figure,
$C_{P}=1+1+2=4\, \mu F$
This combination is in series with $1 \,\mu F$ capacitor.
$\frac{1}{C_{S}}=\frac{1}{4}+\frac{1}{1}=\frac{5}{4}, C_{S}=\frac{4}{5}\, \mu F$
$Q=C V=\frac{4}{5} \times 6=4.8 \,\mu C$
Potential difference across $1\, \mu F$ capacitor,
$V_{1}=\frac{Q}{C}=\frac{4.8 \,\mu C }{1\, \mu F }=4.8 V$
Potential difference across the combination of three capacitors,
$V_{2}=V-V_{1}=6-4.8=1.2 \,V$
