wiz-icon
MyQuestionIcon
MyQuestionIcon
4
You visited us 4 times! Enjoying our articles? Unlock Full Access!
Question

Five capacitors are connected as shown in the figure. Initially, all capacitors are uncharged and S is open. When S is closed, then the potential difference between points M and N is :
[assume steady state to be achieved after S closes]


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

The correct option is B 12 V
After switch S is closed, all the capacitor and both batteries form a closed loop. Let that higher emf battery supplies a charge q in circuit.


Now applying KVL in the circuit in anti-clockwise path,
(All the capacitor will have same q charge )

ΣΔV=0

(q4)(q2)(q4)7(q6)(10q12)+31=0

3q+6q+3q+2q+10q12=24

24q=24×12

q=24×1224=12 μC

Now, considering the path M to N along direction of charge flow;

VM(q4)7(q6)=VN

VMVN=5q6+7

VMVN=(5×1212)+7=12 V

Why this question?Concept: When a higher emf battery ispresent in a closed loop, we assume charge to besupplied in circuit from higher emf battery.Tip: In order to avoid any confusion, its better toapply KVL along direction of charge flow.

flag
Suggest Corrections
thumbs-up
44
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon