Current Electricity Charging Discharging Of Capacitors Question 153
Question: A wire of resistance 10 W is bent to form a circle. P and Q are points on the circumference of the circle dividing it into a quadrant and are connected to a Battery of 3 V and internal resistance 1 W as shown in the figure. The currents in the two parts of the circle are
[Roorkee 1999]
Options:
A) $ \frac{6}{23}A\text{and},\frac{18}{23}A $
B) $ \frac{5}{26}A\text{and},\frac{15}{26}A $
C) $ \frac{4}{25}A\text{and},\frac{12}{25}A $
D) $ \frac{3}{25}A\text{and},\frac{9}{25}A $
Show Answer
Answer:
Correct Answer: A
Solution:
In the following figure Resistance of part PNQ; $ R _{1}=\frac{10}{4}=2.5\Omega $ and Resistance of part PMQ; $ R _{2}=\frac{3}{4}\times 10=7.5\Omega $
$ R _{eq}=\frac{R _{1}R _{2}}{R _{1}+R _{2}}=\frac{2.5\times 7.5}{(2.5+7.5)} $ = $ \frac{15}{8}\Omega $ . Main Current i = $ \frac{3}{\frac{15}{8}+1}=\frac{24}{23}A $ So, i1= $ i\times ( \frac{R _{2}}{R _{1}+R _{2}} )=\frac{24}{23}\times ( \frac{7.5}{2.5+7.5} )=\frac{18}{23}A $ and $ i _{2}=i-i _{1}=\frac{24}{23}-\frac{18}{23}=\frac{6}{23}A $ .
BETA
