A nonalternating periodic waveform is shown below, It’s from factor is
Here, time period is, T = 20 ms; between 10 ms < t < 20 ms, i(t) = 0 A.
Also,
= 50 x 0.01 =0.5 A
The voltage and current in an a.c, circuit are given by
v = 100 sin 314t; i = 10 sin 314t
The average power in the circuit is
∴ P_{avg }= I^{2 }R
Now,
(since v and i in phase)
The expression for instantaneous power dissipated in a resistor R connected across a sinusoidal alternating voltage source represented as v = V_{m} sin ωt is given by
∴ instantaneous power,
The power delivered by the two voltage sources of 6 V and 24 V as shown below will be
Since V_{1} < V_{2}, therefore the current wil! flow from V_{2} to V_{1} as shown below.
Hence, the whole power in the circuit will be supplied by V_{2} only. Total power delivered by the voltage source V_{2} is
P_{T} = I^{2} x (2 + 4) + 6I
= 3^{2 }x 6 + 6 x 3 = 54 + 18
= 72 watts
The power loss in watts in the resistor R shown below is
V_{R} = (10 + 5 V_{R}) x R
Given, V_{R }= 5 V
So, 5 = (10 + 5 x 5) R
or,
∴ Power loss in R is
The root mean square value of the current wave given by i = (50 + 30 sin t) A will be
= 54.30 A
Assertion (A): In an alternating circuit, if the current is (4  j3) amperes and the impressed voltage is (100  j50) volts, the reactive power in the circuit is 100 VA.
Reason (R): True power is the real part of (VI).
V = (100  j50) volt,
I = (4  j3) A
∴ S = VI* = (100  j50)(4  j3)*
= (100  j50)(4 + j3)
= 550 + j100
∴ P = 100 W and Q = 100 VAR
Q is in VAR not in VA. Hence, assertion is not correct.
Also P is the real part of apparent power (S = VI).
In a purely resistive circuit, the average power
P_{av} is______the peak power P_{max}.
Instantaneous power in a purely resistive circuit is
The power consumed in an inductive circuit will be
In a purely inductive circuit, power consumed = Q = VI sin φ. In inductive circuit (having both R and L) power is consumed by resistor R which is P = VI cos φ.
A series circuit containing passive elements has the following current and applied voltage:
The circuit elements
Phase difference between v and i is
∴ power factor = cos φ = cos 75° lag
Since, i lags v therefore, the circuit elements may be resistance and inductance (Z = R + jX_{L}) or it may be resistance, inductance and capacitance Z = R + j(X_{L}  X_{C}), so that p.f. becomes lagging.
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