An alternating voltage 200sin471t is applied to a half wave rectifier which is in series with a resistance of 40 ohm
Answers
Explanation:
An alternating voltage e=200 sin 314 t is applied to a device which offers a resistance of 20 ohm to the flow of current in one direction. What is the value of current?
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The question in my opinion has missing information so one should make basic assumptions:
The Resistance Value of 20 ohm is really an impedence value or else we are missing the real resistance to current flow in an ac circuit. We will assume this is a pure resistance to simplify matters.
We do not know what the impedance of the circuit is when current attempts to flow in the reverse direction. Therefore we can only calculate the value of the current in the forward direction.
The solution now becomes a simplified equation: V=IZ
Therefore I=V/Z which is equal to 200/20 sin314t
I=10sin314t for the value of t where v is positive (depending on the frequency of the waveform supplied)
Explanation:
L.H.S. = \large\rm {K.E. = ML²T^{-2}}K.E.=ML²T
−2
R.H.S. = \large\rm{\frac {[MLT^{-1}]^{2}}{M}}
M
[MLT
−1
]
2
= \large\rm { ML²T^{-2}}ML²T
−2
L.H.S. = R.H.S.
HENCE PROVEN✓