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SS3: PHYSICS - 2ND TERM

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  1. Magnetic Field | Week 1
    4 Topics
  2. Electromagnetic Field
    4 Topics
  3. Electromagnetic Induction
    6 Topics
  4. The Transformer
    5 Topics
  5. Simple A.C Circuit
    4 Topics
  6. Models of the Atom
    2 Topics
  7. Radioactivity
    3 Topics
  8. Half Life
    8 Topics
  9. Energy Quantization
    3 Topics
  10. Photoelectric Effect
    4 Topics
  11. Wave Particle Paradox
    3 Topics



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Lesson 5, Topic 3
In Progress

R-L Circuit | R-C Circuit | L-C Circuit | RLC Circuit

Lesson Progress
0% Complete

Circuit Containing Resistor and Inductor only (R-L)

Screen Shot 2020 10 29 at 7.48.20 PM

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 

Screen Shot 2020 10 29 at 7.49.37 PM

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

Screen Shot 2020 10 29 at 8.04.48 PM

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 

Screen Shot 2020 10 29 at 8.30.02 PM

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)

Screen Shot 2020 10 29 at 8.48.54 PM

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 = \( \frac{1}{\sqrt{LC}} \)  

Screen Shot 2020 10 29 at 8.49.25 PM

Resonance frequency is useful in selecting definite frequency from numerous radio waves. It is applicable in radio tuning and television.

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