Question

# In the circuit shown in figure, emf of the batteries are E1=3 V, E2=2 V, E3=1 V and their internal resistances are R=r1=r2=r3=1 Ω. The potential difference between the points A and B, and the current through branch containing resistor r2 will be respectively

A
2 V, 1 A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3 V, 2 A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2 V, 0 A
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
4 V, 1.5 A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C 2 V, 0 AThe given circuit can be visualized as a parallel combination of three non-ideal batteries, and equivalent battery is connected in series with resistor R. So, equivalent emf and resistance of parallel combination will be Eeq=∑(Er)∑(1r) Putting the value in expression, we get Eeq=31+21+1111+11+11=63 ∴Eeq=2 V Also, equivalent internal resistance, 1req=1r1+1r2+1r3 ⇒1req=1+1+1 ∴req=13 Ω From the equivalent circuit shown in figure, it can be observed that no current will be flown from cell as the circuit is not closed one. ∴VA−VB=Eeq=2 V Now in the branch containing resistance r2, let the current be i2. Apply KVL between points A and B, VA+(0×R)+i2r2−E2=VB ⇒(VA−VB)−E2=−i2r2 ⇒2−2=−i2r2 ∴i2=0 Current through the branch containing r2 is zero. Hence, option (c) is correct.

Suggest Corrections
3
Join BYJU'S Learning Program
Select...
Join BYJU'S Learning Program
Select...