This question was previously asked in

LPSC (ISRO) Technician B (Electronic Mechanic): Previous Year Paper (Held on 4 March 2018)

Option 1 : 10 mA

CT 1: Basic Concepts

18600

10 Questions
10 Marks
6 Mins

__Concept__:

For a series RLC circuit, the net impedance is given by:

Z = R + j (XL - XC)

XL = Inductive Reactance given by:

XL = ωL

XC = Capacitive Reactance given by:

XL = 1/ωC

The magnitude of the impedance is given by:

\(|Z|=\sqrt{R^2+(X_L-X_C)^2}\)

The current flowing across the series RLC circuit will be:

\(I=\frac{V}{|Z|}\)

At resonance, XL = XC, resulting in the net impedance to be minimum. This eventually results in the flow of the maximum current of:

∴\(I=\frac{V}{R}\)

__Application:__

Given:

R = 1kΩ , L = 10μH, C = 1μf and source voltage = 10 V

In a series RLC circuit, at resonance, the impedance of the circuit is equal to the resistance value as Z = R.

∴ Impedance of the circuit at resonance = Z = R = 1kΩ.

\(I=\frac{10}{1 \ k\Omega}\)

I = 10 mA

So, option (1) is correct.

In the series RLC circuit, the current (I) Vs frequency (f) graph for the series resonance circuit is shown below. :-

From above we can conclude:-

- For resonance to occur in any circuit it must have at least one inductor and one capacitor.
- Resonance is the result of oscillations in a circuit as stored energy is passed from the inductor to the capacitor.
- Resonance occurs when XL = XC and the imaginary part of the transfer function is zero.
- At resonance, the impedance of the circuit is equal to the resistance value as Z = R.
- At low frequencies the series circuit is capacitive as XC > XL, this gives the circuit a leading power factor.
- At high frequencies the series circuit is inductive as XL > XC, this gives the circuit a lagging power factor.
- The high value of current at resonance produces very high values of voltage across the inductor and capacitor.
- Because impedance is minimum and current is maximum, series resonance circuits are also called Acceptor Circuits.