A resistance R draws apower P when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes Z the power drawn will be
Answer (1)
7th Aug, 2020
Greeting Aspirant,
The power drawn will be P(r/z)^2. The same cannot be explained here, hence i have solved and uploaded a picture. Still i have tried to give the solution below:
In case of pure resistance circuit,
For pure resistor circuit, power
P=V^2/R⇒V^2=PR
When inductance is connected in series with resistance.
For L−R series circuit, power
P'=V^2/Z *cosθ=PR/Z . R/Z= P(R/Z)^2
The Image link for the same with diagram is below:
https://drive.google.com/file/d/1CbGKSqE4mWGRUZMurwJSR6iduUpHWqX8/view?usp=drivesdk
Best Of Luck For Future.
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