Question

A 100 mH inductor, a 25 µF capacitor and 15 Q resistor are conn

120 V, 50 Hz AC source, calculate current at resonance.​

Answer 1

Explanation:

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Answer 2

The current at resonance will be

The inductance of inductor = 100 mH

                            = 100 × 10⁻³ H

The capacitance of capacitor = 25μF

                                                = 25 × 10⁻⁶ F

Voltage of the source = 120 V

Frequency of the source = 50 Hz

At resonance inductive reactance ($X_{L}$ = ωL) and the capacitive reactance

($X_{C}$= 1/ωC) are same.

So, the impedance, Z = $\sqrt{R^2+(X_{L}-X_{C})^2 }$

                               Z = R  [As $X_{L}$ = $X_{C}$]

Resistance (R) = 15 Ω

So, the current is = 120/15 A

                             = 8 A

The current at resonance is 8 A.