Question

# Four capacitors of capacitance C1=1μF, C2=2μF, C3=3μF and C4=4μF are connected as shown in the figure. Find the potential differences across C3 when switch S is closed.

A
7 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3 V
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
7.5 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2.5 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 3 V C1 & C2 are in parallel, so the equivalent of these two capacitors is given as C′=C1+C2=(1+2)=3μF Similarly, C3 & C4 are in parallel, so C′′=C3+C4=3+4=7μF Since, C′ & C" are in series, so the charge on both capacitors in the circuit will be same. Charge on C′= charge on C′′ Let voltage across C′ and C′′ be V′ and V′′ respectively. ∴C′V′=C′′V′′ V′=C′′C′V′′=73V′′ Also, given that, V′+V′′=10 V 7V′′3+V′′=10 ∴V′′=3 V Therefore, potential difference across C3 & C4 will be 3 volts. So. the right option is (b)

Suggest Corrections
1
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program