A wire of resistance 10 Ω Ω 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 Ω Ω as shown in the figure. The currents in the two parts of the circle are
Text Solution
Verified by ExpertsA
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, i 1 = $i \times \left(\frac{R_2}{R_1 + R_2}\right) = \frac{24}{23} \times \left(\frac{7.5}{2.5 + 7.5}\right) = \frac{18}{23} A$
and $i_{2} = i - i_{1} = \frac{24}{23} - \frac{18}{23} = \frac{6}{23} A$ .
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems