Circuit Containing Resistor and Inductor only (R-L)
The impedance, z = \( \frac{Vrms}{Irms} = \frac{V_0}{I_0} \)
ZRL = \( \scriptsize \sqrt{R^2 \; + \; X_L^2} \)
IO = \( \frac{V_0}{\sqrt{R^2 \; + \; X_L^2}} \)
Voltage leads the current by \( \frac {\pi}{2 } \scriptsize \; or \; 90^{\circ}\)
\( \scriptsize V^2 = V_R{^2} + V_L{^2}\)Phase angle φ is the angle between supply voltage and the current,
Tanφ = \( \frac{V_L}{V_R} = \frac{X_L}{R} \)
Example
An inductor and a resistance are connected in series in an a.c supply. The voltage along the inductor and resistor are 20v and 18v respectively. Calculate the (i) supply voltage (ii) current flowing in the circuit, if impedance is 15Ω (iii) inductance of the coil
Solution
\( \scriptsize V = \sqrt {V_R{^2} + V_L{^2}}\)= \( \scriptsize \sqrt {20{^2} + 18{^2}}\)
= \( \scriptsize \sqrt {400 + 324}\)
= \( \scriptsize \sqrt {724}\)
I = \( \frac{V}{Z} = \frac{20.6}{15} \)
= 1.37A
XL = \( \frac{V_L}{f} =\frac{20}{1.37}\)
= 14.6Ω
XL= 2πfL,
L =\( \frac{X_L}{2 \pi f} \)
\( \frac{14.6}{2 \pi \frac{100}{\pi}} \)= 0.0734H
Circuit Containing Resistor and Capacitor (R-C Circuit)
For R-C circuit,
zRC = \( \frac{Vrms}{Irms} = \frac{V_0}{I_0} \)
ZRC = \( \scriptsize \sqrt{R^2 \; + \; X_C^2} \)
Current in R-c circuit, IO = \( \frac{V_0}{\sqrt{R^2 \; + \; X_c^2}} \)
The current leads the voltage by \( \frac {\pi}{2 } \scriptsize \; or \; 90^{\circ}\)
Tanφ = \( \frac{V_C}{V_R} = \frac{X_C}{R} =\frac{IX_C}{IR} \)
Vtotal = \( \scriptsize V^2 = V_R{^2} + V_C{^2}\)
Example
A 40 F capacitor in series with a 20Ω resistor is connected to a 100v, 50 Hz a.c supply. Draw a circuit diagram for the arrangement and calculate (i) impedance in the circuit (ii) current in the circuit (iii) p.d across the circuit
Solution
C = 40 ,R = 20Ω, f = 50Hz ,Vrms = 100v
(i) ZRC = \( \scriptsize \sqrt{R^2 \; + \; X_C^2} \)
XC = \( \frac{1}{2 \pi f c} \)
XC = \( \frac{1}{2 \pi \; \times \; 50 \; \times \; 40 \times 10^{-6} } \)
XC = 79.56Ω
ZRC = \( \scriptsize \sqrt{20^2 \; + \; 79.58^2} \)
ZRC = 82.05Ω
(ii) I0 = \( \frac{V_{rms}}{\sqrt{R^2 \; + \; X_c^2}} = \frac{V_{rms}}{Z_{RC}} \)
I0 = \( \frac{100}{82.05} \)
= 1.21A
P.d = V = I C
= 1.21 x 79.85
= 89.35V
Circuit Containing Inductor and Capacitor Only (L-C)
The impedance for LC circuit ZLC = \( \frac{Vrms}{Irms} = \frac{V_0}{I_0} \)
ZLC = \( \scriptsize \sqrt{X_L^2 \; – \; X_C^2} = \sqrt{ \left(X_L \; – \; X_C \right)^2} \)
= XL – XC
Current in LC circuit, I0 = \( \frac{V_0}{\sqrt{\left(X_L \; – \; X_C \right)^2}} \)
I0 = \( \frac{V_0}{X_L \; – \; X_C} \)
The inductor voltage leads the capacitor voltage by 180o ,
Effective voltage V = \( \left (\scriptsize V_L – V_C \right)^2 \)
= VL – VC
Example
The frequency of the a.c circuit as shown is \( \frac{600}{\pi} \scriptsize Hz \) Find the reactance of the circuit
L = 0.9H, 2μf = 2 x 10-6 = C
XL= 2πfL,
=\( \scriptsize 2 \pi \; \times \;\normalsize \frac {600}{\pi} \scriptsize \; \times \; 0.9 \)
XL= 1080Ω
Capactive reactance, XC = \( \frac{1}{2 \pi f c} \)
XC = \( \frac{1}{2 \pi \; \times \normalsize \; \frac{600}{\pi} \scriptsize \; \times \; 2 \times 10^{-6} } \)
= 416.7Ω
∴ Circuit reactance, z = \( \scriptsize \sqrt{X_L^2 \; – \; X_C^2} \)
= \( \scriptsize \sqrt{1080^2 \; – \; 416.7^2} \)
= 996.4Ω
Circuit Containing Resistor, Inductor and Capacitor (RLC)
zRC = \( \frac{Vrms}{Irms} = \frac{V_0}{I_0} \)
ZRC = \( \scriptsize \sqrt{R^2 \; + \; X_C^2} \)
Current in R-c circuit, I0 = \( \frac{V_0}{\sqrt{R^2 \; + \; X_c^2}} \)
The current leads the voltage by \( \frac {\pi}{2 } \scriptsize \; or \; 90^{\circ}\)
Tanφ = \( \frac{V_C}{V_R} = \frac{X_C}{R} =\frac{IX_C}{IR} \)
Vtotal = \( \scriptsize V^2 = V_R{^2} + V_C{^2}\)
RLC circuit, impedance, z = \( \frac{Vrms}{Irms} = \frac{V_0}{I_0} \)
Current in RLC circuit , I0 = \( \frac{V_0}{\sqrt{R^2 \; + \; \left( X_L \; – \; X_C \right)^2}} \)
Phase angle φ ,Tanφ = \( \frac{V_L – V_C}{V_R} = \frac{IX_L – IX_C}{V_R} \)
=\( \frac{X_L – X_C}{V_R} \)
Effective voltage, V = \( \scriptsize \sqrt {V_R{^2} + \left ( V_L \; – \;V_C \right)^2}\)
Example
An a.c circuit of e.m.f 24v has a resistance 8Ω connected in series to an inductor of inductive reactance 16Ω and a capacitor of capacitive reactance 6Ω. Find the current flowing in the resistor.
Solution
Current I0 = \( \frac{V_0}{\sqrt{R^2 \; + \; \left( X_L \; – \; X_C \right)^2}} \)
I0 = \( \frac{24}{\sqrt{8^2 \; + \; \left( 16 \; – \; 6 \right)^2}} \)
I0 = \( \frac{24}{\sqrt{64 + 100}} = \frac{24}{\sqrt{164}} \)
I0 = 1.87A
Resonance in R-L-C SERIES CIRCUIT
Resonance occurs in an a.c series circuit when the maximum current is obtained in the circuit. The resonance frequency occurs when the capacitive reactance is equal to the inductive reactance. i.e.
2πfl = \( \frac{1}{2 \pi f c} \)
∴ fo = \( \frac{1}{2 \pi \sqrt{LC}} \)
wo = 2πf
wo = \( \frac{1}{\sqrt{LC}} \)
Resonance frequency is useful in selecting definite frequency from numerous radio waves. It is applicable in radio tuning and television.
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