Current Electricity Charging Discharging Of Capacitors Question 174
Question: Consider the circuits shown in the figure. Both the circuits are taking same current from battery but current through R in the second circuit is $ \frac{1}{10} $ th of current through R in the first circuit. If R is 11 W, the value of R1
Options:
A) 9.9 W
B) 11 W
C) 8.8 W
D) 7.7 W
Show Answer
Answer:
Correct Answer: A
Solution:
In figure (b) current through $ R _{2}=i-\frac{i}{10}=\frac{9i}{10} $ Potential difference across $ R _{2} $ = Potential difference across R
Therefore $ R _{2}\times \frac{9}{10}i=R\times \frac{i}{10} $ i.e. $ R _{2}=\frac{R}{9}=\frac{11}{9}\Omega $
$ R _{eq}=\frac{R _{2}\times R}{(R _{2}+R)}=\frac{\frac{11}{9}\times \frac{11}{1}}{\frac{11}{9}+\frac{11}{1}}=\frac{11}{10}\Omega $ Total circuit resistance $ =\frac{11}{10}+R _{1}=R=11 $
Therefore $ R _{1}=9.9\Omega $
BETA
