Question

A resistance of 10 , an inductor of 10/377 henries, and a capacitor of 0.1/377 farads are all connected in series to a 60-hertz, 100-volt source. Determine the magnitude of the current.

Answer 1

Answer:

The magnitude of the current is 10 A.

Explanation:

Given that,

Resistance, R = 10 ohms

Inductance, $L=\dfrac{10}{377}\ H=0.0265\ H$

Capacitance, $C=\dfrac{0.1}{377}\ F=0.000265\ F$

Voltage, V = 100 V

Frequency, f = 60 Hz  

At resonance condition, the resistance of the circuit is equal to its impedance. It is given by :

$Z=\sqrt{R^2+(2\pi f L-\dfrac{1}{2\pi f C})^2} \\\\Z=\sqrt{10^2+(2\pi \times 60\times 0.0265-\dfrac{1}{2\pi \times 60\times 0.000265})^2}\\\\Z=10\ \Omega$

Let I is the current. It can be calculated using Ohm's law as :

V = I Z

$I=\dfrac{V}{Z}\\\\I=\dfrac{100}{10}\\\\I=10\ A$

So, the magnitude of the current is 10 A. Hence, this is the required solution.