Two cells of same emf E but of different internal resistance r1 and r2 are connected in series with an external resistance R. The potential drop across the first cell is found to be zero. The external resistance R is
A
r1+r2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
r1−r2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
r1r2r1+r2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
√r1r2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Br1−r2 The total current in the circuit is I=2Er1+r2+R ..............................(1).....(In series, E+E=2E) Now the potential drop across first cell is V1=E−Ir1 As per the given condition V1=0 hence, E−Ir1=0 E−(2Er1+r2+R)r1=0 .....................from(1) 2r1Er1+r2+R=E ⇒r1+r2+R=2r1 R=r1−r2