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

The magnitude of the current is 10 A.

Step-by-step explanation:

Given that,

Resistance, R = 10 ohms

Inductance,
`L=(10)/(377)\ H=0.0265\ H`

Capacitance,
`C=(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-(1)/(2\pi f C))^2} \\\\Z=\sqrt{10^2+(2\pi * 60* 0.0265-(1)/(2\pi * 60* 0.000265))^2}\\\\Z=10\ \Omega`

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

V = I Z

`I=(V)/(Z)\\\\I=(100)/(10)\\\\I=10\ A`

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