Published by:
CGP EDU Academic Team
Published on: September 12, 2026
The switch S has been closed for long time and the electric circuit shown carries a steady current. Let C 1 = 3.0 µF, C 2 = 6.0 µF, R 1 = 4.0 kΩ, and R 2 = 7.0 kΩ. The power dissipated in R 2 is 2.8 W.
(i)The power dissipated to the resistor R 1 is-
Correct Matrix Matching
Text Solution
Verified by ExpertsThe correct answer is:
(i) Sol; (ii) Sol; (iii) Sol
Ans.
(i)
Sol. i = 2 × 10 –2 A
i 2 R 1 = (2.0 × 10 –2 ) 2 × 4 × 10 3 = 1.6 W
(ii)
Sol. 
× C 1 = 80 × 3 × 10 –6 = 240 µC
× C 2 = 140 × 6 × 10 –6 = 840 µC
(iii)
Sol. 
= EC 1 = 220 × 3 × 10 –6 = 660 µC
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
Two equal negative charge – q are fixed at the fixed points (0,a) and $(0,-a)$ on the Y-axis. A pos…
An electric line of force in the xy plane is given by equation $x^{2} + y^{2} = 1$ . A particle wit…
A positively charged ball hangs from a silk thread. We put a positive test charge $\alpha_0$ at a p…
If on the concentric hollow spheres of radii r and $R(>r)$ the charge $QED$ is distributed such tha…
Two equal charges of opposite sign separated by a distance constitute an electric dipole of dipol…
A point charge q is placed at a distance a/2 directly above the center of a square of side a. The e…