Current Electricity Charging Discharging Of Capacitors Question 717
Question: Two sources of equal emf are connected to an external resistance R. The identical resistance of the two sources are $ R _{1} $ and $ R _{2} $ $ ( R _{2}>R _{1} ) $ . If the potential difference across the source having internal resistance $ R _{2} $ is zero, then
Options:
A) $ R=R _{2}-R _{1} $
B) $ R= \frac{R _{2}( R _{1}+R _{2} )}{( R _{2}-R _{1} )} $
C) $ R=\frac{R _{1}R _{2}}{( R _{2}-R _{1} )} $
D) $ R=\frac{R _{1}R _{2}}{( R _{1}-R _{2} )} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ I=\frac{2\varepsilon }{R+R _{1}+R _{2}} $ Pot. difference across second cell $ =V=\varepsilon -IR _{2}=0 $
$ \varepsilon =\frac{2\varepsilon }{R+R _{1}+R _{2}}.R _{2}=0 $
$ R+R _{1}+R _{2}-2R _{2}=0 $
$ R+R _{1}-R _{2}=0\text{ } $
Therefore $ R=R _{2}-R _{1} $
BETA
