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: Power-loss expression.
The mean power lost per cycle is \(P=E_{rms}I_{rms}\cos\phi\), the product of rms voltage, rms current and the power factor \(\cos\phi\); all of it appears as heat in \(R\), i.e. \(P=I_{rms}^{2}R\).
Step 2: Draw the voltage triangle.
Place \(V_R=40\) V along the current direction. The reactive voltages act along the perpendicular: net reactive voltage \(=V_L-V_C=50-20=30\) V. The source voltage is the hypotenuse of the right triangle with legs 40 V and 30 V.
Step 3: Hypotenuse gives the emf.
\(E=\sqrt{40^{2}+30^{2}}=\sqrt{1600+900}=\sqrt{2500}=50\) V. (A neat 3-4-5 triangle scaled by 10.)
Step 4: Angle of the triangle.
\(\cos\phi=\dfrac{V_R}{E}=\dfrac{40}{50}=0.8\), so \(\phi=\cos^{-1}(0.8)=37^\circ\). Because the inductive side dominates (\(V_L>V_C\)), the voltage is ahead of the current by this angle.
\[\boxed{E=50\ \text{V},\ \phi\approx 37^\circ}\]