circuit theoremcircuit theorem and analysisCircuit theorem II

With a suitable diagram show the alternating current and voltage in a pure resistance are to phase. Hence, derive an expression for power in a pure a.c resistive circuit.

CIRCUIT THEOREM

1. With a suitable diagram show the alternating current and voltage in a pure resistance are to phase. Hence, derive an expression for power in a pure a.c resistive circuit.

R = resistance
From ohm’s law,
V = IR
VmSinwt = IR
I =(VmsinWt) ÷ R
Recall SinWt = 1
Current(I) is maximum(IM)
IM = Vm/R
Therefore,
Instantaneous current (I) = IMSinWt
Power across the circuit, V = VMSinWt
P = IV
P = IMSinWt × VMSinWt
P = VMIMSin2Wt
P =VmIM (1-cos2Wt)/2

2. A 60hz voltage of 180V (r.m.s) is impressed on 150Ω resistance. Write the time equation for the voltage across the resulting current.


F = 60hz, Vr.m.s = 180v, R = 150Ω
W = 2Πf = 2×3.142×60 = 377rads/sec.
VMAX = root 2 × Vr.m.s = root 2 × 180 = 254.56v
Vt = VmSinwt
Vt = 254.56Sin377t
Time equation for the voltage (Vt ) = 254.56Sin377t
I(t) = It Sinwt
But, maximum current (IM) = VM/R = 254.56/150 = 1.7A\
I(t) = 1.7Sin377t
Time equation for the resulting current (It) = 1.7Sin377t Ampere

Oluwamuyide Peter

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