wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If 100 volts of potential difference is applied between a and b in the circuit shown in the figure. The potential difference between 'c' and 'd' will be (Given Va>Vb).


A

2003 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

1003 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

1003 V
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

50 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C
1003 V
From the circuit given, we can see that the three 6 μF capacitors connected in upper branch are in series.

Now simplifying the circuit and redrawing it,


Since the combination of three 6 μF capacitors are connected in parallel with battery. So,

Sum of potential difference across 6 μF capacitors = potential difference of battery

Since, all the three capacitors are identical, hence

V1=V2=V3=V

V1+V2+V3=100V

3V=100 V

V=1003 V

From the shown figure, direction of flow of positive charge it is clear that Vc>Vd

VcVd=+1003 V

Alternative solution:

Let the three 6 μF capacitors be replaced an equivalent capacitor

1Ceq=16+16+16

Ceq=63=2μF



Charge flow through the 2 μF, q1=100×2=200 μC

The potential difference,

VcVd=(q1)6 μF

VcVd=200 μC6 μF=1003V

Hence, option (c) is correct answer.

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon