The current, I, in an RLC series circuit is given by

I = V/Z = V/√R^{2 }+ (X_{L }– X_{C})^{2}

Where X_{L }= 2πfL and X_{C }= 1/2πfC. The maximum current is obtained in the circuit when the impedance Z, in the above equation is minimum. This happens when

X_{L }= X_{C }or 2πfL = 1/2πfC

**Resonance** is said to occur in a.c. series circuit when the maximum current is obtained from such a circuit.

The frequency at which this resonance occurs is called the resonance frequency (f_{0}). This is the frequency at which X_{L }= X_{C }or 2πf_{0}L = 1/2πf_{0}C.

Hence solving the above equation we obtain that f_{0} is given by

f_{0} = 1/2π√LC

or since ω = 2πf, we can write the condition of resonance as:

ω_{0 }= 1/√LC

The variation of current I, and frequency f, in an RLC series circuit

## Resonance in RLC Series Circuit

https://www.slideshare.net/slideshow/embed_code/key/JnK6fDIzjkpcn4**RLC series circuit simulation at Proteus **from **Wakil Kumar**

**Application of Resonance**

The resonant circuit finds applications in electronics. It is used to tune radios and TVs. Its great advantage is that it responds strongly to one particular frequency. The other frequencies are very little effect.

Hence such a resonant circuit can select one signal of a definite frequency from a jumble of other signals available to it. That resonance frequency corresponds to that of a particular incoming radio signal. When this happens, maximum current is obtained and the distant radio station is loudly and clearly heard.

**Example**

An a.c. voltage of amplitude 2.0 volts is connected to an RLC series circuit. If the resistance in the circuit is 5 ohms, and the inductance and capacitance are 3 mH and 0.05μF respectively, calculate

- the resonance frequency
*f*_{0} - the maximum a.c. current at resonance

Solution

*f*_{0 }= 1/2π√LC

= 1/2π√3 x 10^{-3 }x 5 x 10^{-8} = 1/2π√15 x 10^{-11}

= 1299.545 Hz

= 13 KHz

At resonance Z = R since X_{L }= X_{C}

I_{0} = V_{0}/R = 2/5 = 0.4 Amps

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