## Note on A.C. through an Inductance and Resistance and A.C. through Capacitance and a Resistance

• Note
• Things to remember

#### A.C. through an Inductance and Resistance

Suppose a pure resistance R and a pure inductance L is connected in series to a source of alternating e.m.f. shown in a figure. Let E be the r.m.s value of applied alternating e.m.f. and I be the r.m.s value of current flowing in the circuit.

The potential difference across the inductor, $$V_L = IX_L$$ (leads current I by an angle of p/2).

The potential difference across R, $$V_R = I.R.$$ (in phase with the current).

Since VR and I are in phase so VR is represented by OA in the direction of I. the current lags behind the potential difference VL by angle of $$\pi /2$$ so VL is represented by OB perpendicular to the direction of I. So the resultant of VR and VL is given by OH. The magnitude of OH is given by

\begin{align*} OH &= \sqrt {OA^2 + OB^2} = \sqrt {V_R^2 + V_L^2} \\ \text {or,} \: E &= \sqrt {I^2R^2 + I^2X_L^2} \\ \text {or,} \: E &= I\sqrt {R^2 + X_L^2} \\ \text {or,} \: \frac EI &= \sqrt {R^2 + X_L^2} \\\end{align*}

But $$\frac EI = z$$, is effective opposition of L-R circuit to a.c. called impedance of LR circuit. The impedance of L-R circuit is given by

\begin{align*} Z &= \sqrt {R^2 + X_L^2} \\ \text {Again,} \\ I &= \frac {E}{Z} = \frac {E}{\sqrt {R^2 + X_L^2}} \\ \therefore I &= \frac {E}{\sqrt {R^2 + (L\omega )^2}} \:\:\: [ \therefore X_L = L\omega ] \\ \text {Let} \:\theta \: \text {be the angle between E and I, so from figure, } \\ \text {we have}, \\ \tan \theta &= \frac {V_L}{V_R} = \frac {IX_L}{IR} \\ \text {or,} \: \tan \: \theta &= \frac {X_L}{R} = \frac {L\omega }{R} \\ \therefore \tan \: \theta &= \frac {L\omega }{R} \\ \end{align*}

If the values of X_L and R are known as θ can be calculated. Current lags behind applied voltage or e.m.f.

#### A.C. through Capacitance and a Resistance

Let a pure resistance R and an ideal capacitor of capacitance C be connected in series to the source of alternating e.m.f shown in a figure. Let E be the r.m.s value of applied alternating e.m.f and I be the r.m.s value of current flowing in the circuit. The potential difference across R, $$V_R = IR$$ (in phase with the current).

The potential difference across C is $$V_C = I X_c$$ (lags behind the current by I by angle $$\pi /2$$.

Since VR and VC is given by OH. The magnitude of OH is given by

\begin{align*} OH &= \sqrt {OA^2 + OB^2} \\ \text {or,} \: E &= \sqrt {V_R^2 + V_C^2} = \sqrt {(IR)^2 + (IX_c)^2} \\ \text {or,} \: E &= I \sqrt {R^2 + X_c^2} \\ \text {or,} \: I &= \frac {E}{\sqrt {R^2 + X_c^2}} \\ \text {or,} \: \frac EI &= \sqrt {R^2 + X_c^2} \\ \end{align*}

But E/I = z, is the effective opposition of C-R circuit to a.c. called impedance of C-R circuit. The impedance of C-R circuit is given by

\begin{align*} Z &= \sqrt {R^2 + X_c^2 } \\ I &= \frac EZ = \frac {E}{\sqrt {R^2 + X_c^2}} \\ \therefore I &= \frac {E}{R^2 + ( \frac {1}{\omega c})^2} \\ \text {Let} \: \theta \: \text {be the angle between E and I} \\ \text { So from figure, we have,} \\ \tan \: \theta &= \frac {-V_c}{V_R} = \frac {-IX_c}{IR} \\ \tan \theta &= \frac {X_c}{R} \\ \end{align*}

Since current is taken as the reference phasor, negative phase angle implies that voltage lags behind the current. It is same as current leads the voltage or emf.

Reference

Manu Kumar Khatry, Manoj Kumar Thapa,et al. Bhesha Raj Adhikari, Arjun Kumar Gautam, Parashu Ram Poudel.Principle of Physics. Kathmandu: Ayam publication PVT LTD, 2010.

S.K. Gautam, J.M. Pradhan. A text Book of Physics. Kathmandu: Surya Publication, 2003.

The effective opposition of L-R circuit to a.c. called impedance of LR circuit.

The effective opposition of C-R circuit to a.c. called impedance of C-R circuit.

Current lags behind applied voltage or e.m.f in A.C. through an inductance and resistance.

Applied voltage or e.m.f lags behind current in A.C. through a capacitance and resistance.

.

0%