Current Electricity Charging Discharging Of Capacitors Question 234
Question: A resistance of $ 4,\Omega $ and a wire of length 5 metres and resistance $ 5,\Omega $ are joined in series and connected to a cell of e.m.f. 10 V and internal resistance $ 1,\Omega $ . A parallel combination of two identical cells is balanced across 300 cm of the wire. The e.m.f. E of each cell is
[MP PMT 1997]
Options:
A) 1.5 V
B) 3.0 V
C) 0.67 V
D) 1.33 V
Show Answer
Answer:
Correct Answer: B
Solution:
$ E=x,l=\frac{V}{l}=\frac{iR}{L}\times l $
Therefore $ E=\frac{e}{(R+R _{h}+r)}\times \frac{R}{L}\times l $
Therefore $ E=\frac{10}{(5+4+1)}\times \frac{5}{5}\times 3=3,V $
BETA
