Current Electricity Charging Discharging Of Capacitors Question 597
Question: A cell whose e.m.f. is 2 V and internal resistance is $ 0.1,\Omega $ , is connected with a resistance of $ 3.9,\Omega $ . The voltage across the cell terminal will be
[CPMT 1990; MP PET 1993; CBSE PMT 1999; AFMC 1999; Pb. PMT 2000; AIIMS 2001]
Options:
A) $ 0.50,V $
B) $ 1.90,V $
C) $ 1.95,V $
D) $ 2.00,V $
Show Answer
Answer:
Correct Answer: C
Solution:
The voltage across cell terminal will be given by $ =\frac{E}{R+r}\times R $
$ =\frac{2}{(3.9+0.1)}\times 3.9=1.95V $
BETA
