Question

Give formula for power loss in an alternating circuit. Find the e.m.f. of source in the circuit and the phase difference between resultant potential and the current flowing in the circuit. (Series R-L-C driven by an AC source with \(V_R=40\) V across the resistor, \(V_L=50\) V across the inductor and \(V_C=20\) V across the capacitor.)

Hint:

Power = E_rms I_rms cos(phi). Add the series voltages as phasors: E = sqrt(V_R^2 + (V_L - V_C)^2) and tan(phi) = (V_L - V_C)/V_R.

Answer 1

Step 1: Formula for power loss in an AC circuit.
Average power dissipated is
\[P=E_{rms}\,I_{rms}\cos\phi\]where \(\cos\phi\) is the power factor. Only the resistor consumes power, so equivalently \(P=I_{rms}^2 R\). (Pure L and C dissipate no power.)
Step 2: Combine the voltages (they are not in phase).
In a series R-L-C circuit \(V_R\) is in phase with the current, \(V_L\) leads by \(90^\circ\) and \(V_C\) lags by \(90^\circ\). Hence the source e.m.f. is the phasor sum:
\[E=\sqrt{V_R^2+(V_L-V_C)^2}\]Step 3: Substitute the given values.
\(V_R=40\) V, \(V_L=50\) V, \(V_C=20\) V.
\[E=\sqrt{40^2+(50-20)^2}=\sqrt{1600+900}=\sqrt{2500}=50\ \text{V}\]Step 4: Phase difference.
\[\tan\phi=\frac{V_L-V_C}{V_R}=\frac{50-20}{40}=\frac{30}{40}=0.75\]\[\phi=\tan^{-1}(0.75)=36.87^\circ\approx 37^\circ\]Since \(V_L>V_C\), the circuit is inductive, so the applied voltage leads the current by about \(37^\circ\).
\[\boxed{E=50\ \text{V},\ \phi\approx 37^\circ\ (\text{voltage leads current})}\]