1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Cells A and B and a galvanometer G are connected to a slide wire OS by two sliding contacts C and D as shown in figure. The slide wire is 100 cm long and has a resistance of 12 Ω. With OD=75 cm, the galvanometer gives no deflection when OC is 50 cm. If D is moved to touch the end of wire S, the value of OC for which the galvanometer shows no deflection is 62.5 cm. The e.m.f. of cell B is 1.0 V. Calculate the internal resistance & e.m.f. of cell A.

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

## The correct option is A 3 Ω and 2 VGiven: EB=1 V Let EA and r be the e.m.f. and internal resistance of cell A, respectively. When OD=75 cm and OC=50 cm Resistance of wire OC and OD will be ROC=(12100)×50=6 Ω ROD=(12100)×75=9 Ω Since, EB=1 V is balanced across 50 cm, so potential gradient of wire will be K=150 V/cm Therefore, voltage drop across the wire OD of length 75 cm will be VOD=(1/50)×75=1.5 V So, (EA9+r)×9=1.5 ........(1) Similarly, When OD=100 cm and OC=62.5 cm Resistance of wire OD will be ROD=12 Ω Since, EB=1 V is balanced across 62.5 cm, so potential gradient of wire will be K=162.5 V/cm Therefore, voltage drop across the wire OD of length 75 cm will be VOD=(1/62.5)×100=1.6 V So, (EA12+r)×12=1.6 ........(2) On solving eqs. (1) & (2), we get r=3 Ω and EA=2 V Hence, option (a) is correct. Key concept: Principle: The potentiometer is based upon the principle that when a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.

Suggest Corrections
1
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program