LR, CR, LCR Circuits

SERIES L–R CIRCUIT

As we know potential difference across a resistance in ac is in phase with current and it leads in phase by 90° with current across the inductor.

SERIES C–R CIRCUIT

Potential difference across a capacitor in ac lags in phase by 90° with the current in the circuit.

Suppose in phasor diagram current is taken along positive x-direction. Then VR is also along positive x-direction but VC is along negative y-direction. So we can write.

V = V_R – jV_C = iR – j(iX_C)

=iR – j$\left(\frac{i}{\omega C}\right)$ =iZ

Here, impedence is, Z = R - j$\left(\frac{1}{\omega C}\right)$

The medulus of impedance is,

$\left|Z\right|=\sqrt{R^2+{\left(\frac{1}{\omega C}\right)}^2}$

and the potential difference lags the current by an angle,

$\varphi ={\tan }^{-1}\left|\frac{V_C}{V_R}\right|={\tan }^{-1}\frac{X_C}{R}$ =

${\tan }^{-1}\left(\frac{1/\omega C}{R}\right)$ or

$\varphi ={\tan }^{-1}\left(\frac{1}{\omega CR}\right)$

SERIES L–C–R CIRCUIT

Potential difference across an inductor leads the current by 90° in phase while that across a capacitor, it lags in phase by 90°.

Suppose in a phasor diagram current is taken along positive x-direction. Then V_R is along positive x-direction, Then V_R is along positive x-direction, V_L along positive y-direction and V_C along negative y-direction.

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

So, we can write, V = V_R + jV_L – jV_C = iR + j(iX_L) – j(iX_C)

= iR + j [ i (X_L – X_C)] = iZ

Here impedence is, Z = R + j(X_L – X_C) = R + j$\left(\omega L-\frac{1}{\omega C}\right)$

The modulus of impedence is, $\left|Z\right|=\sqrt{R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2}$

And the potential difference leads the current by an angle,

$\varphi ={\tan }^{-1}\left|\frac{V_L-V_C}{V_R}\right|={\tan }^{-1}\left(\frac{X_L-X_C}{R}\right)$ or

$\varphi ={\tan }^{-1}\left(\frac{\omega L-\frac{1}{\omega C}}{R}\right)$

Illustration : An alternating emf 200 virtual volts at 50 Hz is connected to a circuit of resistance 1$\Omega$ and inductance 0.01 H. What is the phase difference between the current and the emf in the circuit. Also find the virtual current in the circuit.

Solution: In case of an ac, the voltage leads the current in phase by an angle,

$\varphi ={\tan }^{-1}\left(\frac{X_L}{R}\right)$

Here, X_L = $\omega$L = (2$\pi$fL) = (2$\pi$) (50) (0.01) = $\pi$$\Omega$

and R = 1$\Omega$

$\therefore$ $\varphi$ = tan^-1 ($\pi$) » 72.3°

Further, i_rms = $\frac{V_{rms}}{\left|Z\right|}$ $=\frac{V_{rms}}{\sqrt{R^2+X_L^2}}$

Substituting the values we have,

i_rms $=\frac{200}{\sqrt{{(1)}^2+{(\pi )}^2}}$ = 60.67 amp.

CIRCUIT ELEMENTS WITH AC