2
\$\begingroup\$

enter image description here

enter image description here

I am getting wrong answer for the above problem. My book says R should be 23ohms. Please guide me where I am doing the mistake.

\$\endgroup\$
8
  • 4
    \$\begingroup\$ 230V is already in RMS, passing through the bridge still 230V... so 230V/10A = 23 ohms \$\endgroup\$ Jan 16, 2018 at 8:48
  • 1
    \$\begingroup\$ If you replace the current source with a resistive load then the current and the voltage have the same waveform (V = I . R). Thus the peak value of the current will be \$10 \sqrt2\$. So \$R=V_{pk}/I_{pk}=230 \sqrt2 / 10 \sqrt2 = 23 \Omega.\$ \$\endgroup\$ Jan 16, 2018 at 8:51
  • \$\begingroup\$ 230V * sqrt(2) is the peak voltage, but you need Current in RMS \$\endgroup\$ Jan 16, 2018 at 8:51
  • 1
    \$\begingroup\$ Sir, what is wrong in my solution ? Shouldn't the Pavg be Irms^2R ? \$\endgroup\$ Jan 16, 2018 at 8:52
  • 3
    \$\begingroup\$ You supposed that the current is pure DC (from what I've seen in your solution) so the mean (avg) value is the same as RMS. But it's not. It's avg value will be less then its RMS value. \$\endgroup\$ Jan 16, 2018 at 8:53

1 Answer 1

1
\$\begingroup\$

230V is already in RMS, passing through the bridge still 230V... so 230V/10A = 23 ohms – Dr Yunke Jan 16 at 8:48

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.