Question

a 100 mH inductor , a 20 uF capacitor and a 10 ohm resistor are connected in series to a 100 V , 50 Hz ac source . calculate (i) impedance of the circuit at resonance (ii) current at resonance (iii) resonant frequency

Answer 1

(i) The impedance of the circuit at resonance is 10 Ω.

(ii) The current at resonance is 10 A.

(iii) The resonant frequency is 707 Hz.

Given:

Inductance = L = 100 mH

Capacitance = C = 20 μF = $\bold{20\times 10^-6 \ C}$

Resistance = R = 10 ohm

Voltage = V = 100 V

Frequency = ω = 50 Hz

To find:

(i) impedance of the circuit at resonance = ?

(ii) current at resonance = ?

(iii) resonant frequency = ?

Formula used:

Resonance frequency:

$\bold{\omega=\frac{1}{\sqrt{LC}}}$

Resonance impedance = Resistive value

Resonance current:

$\bold{i=\frac{V}{R}}$

Solution:

Resonance frequency:

$\bold{\omega=\frac{1}{\sqrt{LC}}}$

$\bold{\omega=\frac{1}{\sqrt{0.1\times20\times 10^-6}}}$

$\bold{\omega = 707 \ Hz}$

Resonance impedance:

R = 10 ohm

Resonance current:

$\bold{i=\frac{V}{R}}$

$\bold{i=\frac{100}{10} = 10 \ A}$

Answer 2

Answer:

Given:

Resistance of the resistor, R = 100 Ω

Supply voltage, V = 220 V

Frequency, ν = 50 Hz

(a) The rms value of current in the circuit is given as

I= (V/R)

= (220)/ (100)

= 2.20 A

(b) The net power consumed over a full cycle is given as:

P = VI

= 220 × 2.2 = 484 W

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