Concept Notes on L-R or L-C Circuit

By Prashant Kumar|Updated : October 3rd, 2018

AC Circuits, LCR Circuits

Here circuits are expressed in form of phasor diagrams just for convenience in addition, subtraction of sinusoidal quantities which cannot be solved by simple algebra. The quantities represented in phasor diagrams are not vectors but are solved as vectors.

LCR Circuits

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SERIES L–R CIRCUIT

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VL= Ldi/dt

= L d(imsin wt)/dt

= Limw coswt = Limwsin (π/2+wt) (V leads by π/2)

XL=Lw called inductive reactance (just like resistance, unit ohm)

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SERIES C–R CIRCUIT

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XC = 1/wC called capacitive reactance (just like resistance, unit ohm).

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SERIES L–C–R CIRCUIT

Results can also be directly deduced by phasor diagrams.

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Using the figure: Vnet= Vm sin(wt+f)

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V leads I by an angle of ∅

The modulus of impeadence is,

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Example:

An alternating emf 200 virtual volts at 50 Hz is connected to a circuit of resistance 1Ωand 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,

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Here, XL = ωL = (2ΠfL) = (2Π) (50) (0.01) = ΠΩ

and R = 1Ω

∴Φ = tan-1 (Π) ≈ 72.3°

Further,

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Substituting the values we have,

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Example:

Find the voltage across the various elements, i.e., resistance, capacitance and inductance which are in series and having values 1000Ω, 1μF nd 2 henry respectively. Given emf as, V = √100 sin 1000 t volt

Solution:

The rms value of voltage across the source,

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The current will be same everywhere in the circuit, therefore,

P.D across resistor VR = irmsR = 0.0707 ∗1000 = 70.7 volt

P.D across inductor VL = irmsXL = 0.0707 ∗1000 ∗ 2 = 141.4 volt and

P.D across capacitor VC = irmsXC = 0.0707 ∗ = 70.7 volt

Note:

  1. The rms voltages do not add directly as,
    VR + VL + VC = 282.8 volt

Which is not the source voltage 100 volts. The reason is that these voltages are not in phase and can be added by vector or by phasor algebra.

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