Two cells with the same emf E and different internal resistances r1 and r2 are connected in series to an external resistance R. The value of R so that the potential difference across the first cell be zero, is
A
√r1r2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
r1+r2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
r1−r2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
r1+r22
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Br1−r2 The two cells have potential E each and respective internal resistance r1 and r2.
Let the current flowing in the circuit be i.
Using Kirchhoff's voltage law:
⇒−E+ir1+iR−E+ir2=0
⇒i=2E(R+r1+r2)
Potential across first cell, VAB=ir1−E=2Er1(R+r1+r2)−E