Current Electricity Charging Discharging Of Capacitors Question 301
Resistance of 100 cm long potentiometer wire is 10 Ω, it is connected to a battery (2 volt) and a resistance R in series. A source of 10 mV gives null point at 40 cm length, then external resistance R is
[MP PMT 2003]
Options:
A) 490 W
B) 790 W
C) 590 W
D) 990 W
Show Answer
Answer:
Correct Answer: B
Solution:
$ E=\frac{e}{(R+R _{h}+r)}.\frac{R}{L}\times l $
$ \Rightarrow 10\times {{10}^{-3}}=\frac{2}{(10+R+0)}\times \frac{10}{1}\times 0.4 $
Therefore R = 790W
BETA
