A single phase kWhr meter makes 500 revolutions per kWhr. It is found on testing that it is making 40 revolutions in 58.1 seconds at 5kW load. Find out the percentage of error.
Energy consumed in 58.1 seconds=\(\frac{5\ast58.1}{3600}\) kWh=0.08069kWh
Energy consumption registered=\(\frac{Number\ of\ revolution\ made}{meter\ constant\ in\ revolution/kWh}\)
\( =\frac{40}{500}=0.08kWh\)
Percentage error =\(\frac{Actual\ registration-true\ energy\ consumption}{True\ energy\ consumption}\ast100\)
\(=\frac{0.08-0.08069}{0.08069}*100=-0.8555% \)
How can potentiometer be used for
Explain the difference between Dynamometer type wattmeter and induction type wattmeter.
What is Carson’s rule? Explain it.
What is scattering Parameters? Why is it used in microwave Network?
Explain briefly the VSB-SC modulation
