Question

An a.c. source of voltage V = Vm sin ωt is connected, one-by-one, to three circuit elements X, Y and Z. It is observed that the current flowing in them,
(i) is in phase with applied voltage for element X.
(ii) lags the applied voltage, in phase, by $π$ /2 for element Y.
(iii) leads the applied voltage, in phase, by $π$ /2 for element Z. Identify the three circuit elements.
Find an expression for the (a) current flowing in the circuit, (b) net impedance of the circuit, when the same a.c. source is connected across a series combination of the elements X, Y and Z. (c) If the frequency of the applied voltage is varied, set up the condition of frequency when the current amplitude in the circuit is maximum. Write the expression for this current amplitude.

Solution

Given, an a.c. source of voltage V = Vm sin ωt is connected to three circuit elements X, Y and Z one by one.
(a)
(i) Circuit element X is resistance R.
(ii) Circuit element Y is capacitance C.
(iii) Circuit element Z is capacitance L.

(a) Expression for current flowing in the circuit is,
I = I_m sin

ωt

(For R)
I = I_m sin

(ωt - π / 2)

(For L)
I = I_m sin (

ωt + π / 2

) (For C)

(b) Let a resistance R, capacitance C and inductance L be connected in series to a source of alternating e.m.f., as shown in figure (a).

Since R, L and C are in series, therefore, current at any instant through three elements has the same amplitude and phase.

Let it be given as I = I₀ sin

ωt

.
However, voltage across each element bears a different phase relationship with the current. Now,

(i) The maximum voltage across R is

\vec{V_{R}} = \vec{I_{0}} R

In Fig.(b), current phasor

\vec{I_{0}} .

is represented along OX.

Fig.(b)

As

\vec{V_{R}}

is in phase with current, it is represented by the vector

\vec{O} A,

along OX.

(ii) The maximum voltage across L is

\vec{V_{L}} = \vec{I_{0}} X_{L}

.

As voltage across the inductor leads the current by 90°, it is represented by

\vec{OB}

along OY, 90° ahead of

\vec{I_{0}} .

(iii) The maximum voltage across C is

\vec{V_{C}} = \vec{I_{0}} X_{C}

As voltage across the capacitor lags behind the alternating current by 90°, it is represented by

\vec{OC}

rotated clockwise through 90° from the direction of

\vec{I_{0}} .

\vec{OC}

is along OY' is along OY'.
As the voltage across L and C have a phase difference of 180°, the net reactive voltage is

(\vec{V_{L}} - \vec{V_{C}}),

assuming that

\vec{V_{L}} > \vec{V_{C}} .

In figures (a) and (b), it is represented by

\vec{OB'} .

The resultant of the resultant of

\vec{OA}

and

\vec{OB'}

is the diagonal

\vec{OK}

of the rectangle OAKB'.
Hence the vector sum of

\vec{V_{R}}, \vec{V_{L}} and \vec{V_{C}}

is phasor

\vec{E_{0}}

represented by

\vec{OK},

making an angle

ϕ

with current phasor

\vec{I_{0}} .

OK = \sqrt{{OA}^{2} + OB'^{2}}

∴

E_{0} = \sqrt{V_{R}^{2} + {(V_{L} - V_{C})}^{2}} = \sqrt{(I_{0} R)^{2} + (I_{0} X_{L} - I_{0} X_{C})^{2}}

E_{0} = I_{0} \sqrt{R^{2} + (X_{L} - X_{C})^{2}}

The total effective resistance of RLC circuit is called Impedance of the circuit. It is represented by Z.

(c) When the current amplitude in the circuit is maximum then X_L = X_c.

2 {πf}_{0} L = \frac{1}{2 {πf}_{0} C}

f_{0} = \frac{1}{2 π \sqrt{LC}}

where f₀ is called the resonant frequency.

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