Question:

A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is -

Show Hint

Find R from P = I^2 R, then use P' = V^2/R.
Updated On: Nov 5, 2026
  • \(11\times10^{-5}\) W
  • \(11\times10^{-3}\) W
  • \(11\times10^{5}\) W
  • \(11\times10^{-4}\) W
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

\(R=\frac{P}{I^2}=\frac{4.4}{(2\times10^{-3})^2}=\frac{4.4}{4\times10^{-6}}=1.1\times10^6\ \Omega\).
\(P'=\frac{V^2}{R}=\frac{121}{1.1\times10^6}=1.1\times10^{-4}\) W \(=11\times10^{-5}\) W.
Answer: (1) \(11\times10^{-5}\) W
Was this answer helpful?
0
0

Top JEE Main Physics Questions

View More Questions