find the potential difference across ab in the capacitor Share with your friends Share 0 Anshu Agrawal answered this Dear student, The above circuit is simplified by replacing C2 ,C3 ,C4 series capacitance by its equivalent 1C eq = 1C2+1C3 +1C4 ∴C eq =2μF Next applying KVL(Kirchoffs Voltage Law) To Loop I +V−V1=0 ∴V− C1Q1 =0 ... 1 To Loop II −V1+Veq =0 ∴ C1Q1 +C eqQ eq =0 V=30V ,C1 =6μF , C eq =2μF solving 1 and 2 we get ∴ C1Q1 =C eqQ eq V=30 Volt ∴Q=30×2μF=60μC Consider the series combination of C2 ,C3 ,C4 we have Q2=Q3=Q4=Q Therefore between points A and B the voltage drop is given as V3=C3Q3 = 6μF60μC =10V Thus the potential difference between points A and B is 10V Regards 0 View Full Answer