Current Electricity Charging Discharging Of Capacitors Question 654
Question: A cell of constant e.m.f. first connected to a resistance $ R _{1} $ and then connected to a resistance $ R _{2} $ . If power delivered in both cases is then the internal resistance of the cell is
[Orissa JEE 2005]
Options:
A) $ \sqrt{R _{1}R _{2}} $
B) $ \sqrt{\frac{R _{1}}{R _{2}}} $
C) $ \frac{R _{1}-R _{2}}{2} $
D) $ \frac{R _{1}+R _{2}}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
Power dissipated $ =i^{2}R={{( \frac{E}{R+r} )}^{2}}R $ \ $ {{( \frac{E}{R _{1}+r} )}^{2}}R _{1}={{( \frac{E}{R _{2}+r} )}^{2}}R _{2} $
Therefore $ R _{1}(R _{2}^{2}+r _{2}+2R _{2}r)=R _{2}(R _{1}^{2}+r^{2}+2R _{1}r) $
Therefore $ R _{2}^{2}R _{1}+R _{1}r^{2}+2R _{2}r=R _{1}^{2}R _{2}+R _{2}r^{2}+2R _{1}R _{2}r $
Therefore $ (R _{1}-R _{2})r^{2}=(R _{1}-R _{2})r^{2}=(R _{1}-R _{2})R _{1}R _{2} $
Therefore $ r=\sqrt{R _{1}R _{2}} $
BETA
