Notes on A.C. through an Inductance, Capacitance and Resistance | Grade 12 > Physics > Alternating Currents | KULLABS.COM

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### A.C. through an Inductance, Capacitance and a Resistance

Let a pure resistance R, a pure inductance L and an ideal capacitor of capacitance C be connected in a series to a source of alternating e.m.f. shown in a figure. As R, L and C are in series, current at any instant through the three elements has the same amplitude and phase. However, a voltage across each element bears different phase relationship with the current. Let E be the r.m.s. the value of the applied alternating e.m.f. to LCR circuit and I be the r.m.s value of current flowing through all the circuit elements.

#### Points to be noted are:

1. The potential difference across inductor, $$V_L = IX_L$$
(leads current I by an angle of $$\pi /2$$)
2. The potential difference across C, $$V_c = IX_c$$
(leads behind the current I by an angle of $$\pi /2$$)
3. The potential difference across R, $$V_R = IR$$
(in phase with the current)
4. Since VR and I are in phase so VR is represented by OA in the direction of I as shown in the figure.
5. The current lags behind the potential difference VL by an angle of $$\pi /2$$, so VL is represented by OB perpendicular to the direction of I.
6. The current leads the potential difference VC by an angle of $$\pi /2$$ so Vc is represented by OF perpendicular to the direction of I.
7. Since VL and VC are in opposite phase, so their resultant $$(V_L – V_C)$$ is represented by OD (Here VL > VC).

The resultant of VR and $$(V_L – V_C)$$ is given by OH. The magnitude of OH is given by

\begin{align*} OH &= \sqrt {(OA)^2 + (OD)^2} = \sqrt {V_R^2 + (V_L – V_C)^2} \\ \text {or,} \: E &= \sqrt {I^2R^2 + (IX_L – IX_C)^2} \\ &= I\sqrt {R^2 + (IX_L – IX_C)^2} \\ \text {or,} \: \frac EI &= \sqrt {R^2 +(X_L – X_C)^2} \\\end{align*} But$$\: E/I = Z$$ is the effective opposition of LCR circuit to A.C. called impedance of the circuit. \begin{align*} \text {So, we have} \\ \ Z &= \frac EI \\ &= \sqrt {R^2 +(X_L – X_C)^2} \dots (i) \\ \text {Let} \: \theta \: \text { be the phase angle between E and I,} \\ \text {so from figure, we have} \\ \tan \theta &= \frac {AH}{OA} \\ &= \frac {V_L – V_C}{V_R} \\ \frac {IX_L – I X_C}{IR} &= \frac {X_L – X_C}{R} \\ \therefore \tan \theta &= \frac {\left (L\omega - \frac {1}{C\omega } \right )}{R} \\ \end{align*}

The equation (i) is the general relation for impedance.

1. When $$X_L = X_C$$, then $$\tan \theta = 0$$ or $$\theta = O^o$$. Hence voltage and current are in same phase.
2. When $$X_L > X_C$$, then $$\tan \theta$$ or $$\theta$$ is positive. Hence voltage leads the current by phase angle $$\theta$$. The a.c. circuit is inductance dominated circuit.
3. When $$X_L < X_C$$, then $$\tan \theta$$ or $$\theta$$ is negative. Hence current leads the voltage by phase angle $$\theta$$. The a.c. circuit is capacitor dominated circuit.

The total effective opposition offered by LCR circuit to alternating current is known as impedance and is denoted by z. The reciprocal of the impedance of a circuit is known as an admittance of the circuit.

$$\text {i.e.} \: \: \: \text {Admitance} \:(A) = \frac 1Z$$

Unit of impedance (Z) of the circuit is ohm.

Unit of admittance of the circuit is ohm-1 i.e. mho or siemen.

Reference

Manu Kumar Khatry, Manoj Kumar Thapa,et al.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 total effective opposition offered by LCR circuit to alternating current is known as impedance.

The reciprocal of the impedance of a circuit is known as an admittance of the circuit.

If the a.c. the circuit is inductance dominated circuit then  voltage leads the current by the phase angle θ.

If the a.c. the circuit is capacitor dominated circuit then  voltage lags the current by the phase angle θ.

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