Question

A resistor of $40\Omega$, an inductor of $5/\pi H$ and a capacitor of $50/\pi \mu F$ are connected in series across a source of alternating voltage of $140\sin 100\pi t$ Volts. Find the r.m.s voltage across the resistor, the inductor and the capacitor. Is algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox

Answer 1

Given

$E=140\sin 140\pi t$

Comparing with $V=V_0\sin \omega t$

$V_0=140V,\omega =100\pi \Rightarrow 2\pi n=100\pi$

$n=50Hz$

$V_{rms}=\frac{V_0}{\sqrt{2}}=\frac{140}{1.4}=100V$

Inductive reactance $X_L=\omega L=100\pi \times \frac{5}{\pi }=500\Omega$

Capacitive reactance, $X_C=\frac{1}{\omega C}=\frac{1}{100\pi \times \left(\frac{50}{\pi }\right)\times {10}^{-6}}=200\Omega$

Impedeance $Z=\sqrt{R^2+(X_L-X_C{)}^2}$

$Z=\sqrt{{400}^2+(500-200{)}^2}$

$=500\Omega$

rms current in circuit, $I=\frac{V_{rms}}{Z}=\frac{100}{500}=0.2A$

Voltage across resistor, $V_R=RI=400\times 0.2=80V$

Voltage across inductor, $V_L=X_{L}I=500\times 0.2=100V$

Voltage across capacitor, $V_C=X_{C}I=200\times 0.2=40V$

Algebric sum of voltages$=80+100-40=160V$

This is because $VR$ and $(V_L-V_C)$ are not in phase but $(V_l-V_C)$ leads $V_R$ by an angle $\frac{\pi }{2}$

$V=\sqrt{V_R^2+(V_L-V_C{)}^2}=\sqrt{{80}^2+{60}^2}=100V$

This resolves the paradox.