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
14V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
12V
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
10V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
15V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B12V 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=12μC
Now, considering the path M to N along direction of charge flow;
VM−(q4)−7−(q6)=VN
⇒VM−VN=5q6+7
∴VM−VN=(5×1212)+7=12V
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.