Question

An AC circuit is composed of a serial connection of a resistor with resistance 50 ,a coil with inductance 0.3 H and a capacitor with
capacitance 15 pF. The circuit is connected to an AC voltage source with amplitude 25 V and frequency 50 Hz. Determine the amplitude
of electric current in the circuit and a phase difference between the voltage and the current.​

Answer 1

Answer:

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Answer 2

Explanation:

• So resistor with resistance R = 50 ohms
• Inductance of the coil L = 0.3 Henry
• Capacitor of capacitance C = 15 µF = 15 x 10^-6 F
• Amplitude of AC voltage Am = 25 V
• Frequency f = 50 Hz
• Now we need to find amplitude and phase difference between voltage and current
• So we have I = Am / ѴR^2 + (ωL – 1/ωC)^2
•               = 25 / Ѵ50^2 + (2Π x 50 x 0.3 – 1 / 2Π x 50 x 15 x 10^-6)^2 (since ω = 2ΠR)
•              =  25 / Ѵ2500 + (94.2 – 1/4710 x 10^-6)
•                = 25 / Ѵ2500 + (118.114)^2
•                 = 25 / Ѵ2500 + 13950.91
•                  = 25 / Ѵ16450.91
•                   = 25 / 128.26
•                      = 0.19
•                       = 0.2 Amps
• Now to find the phase difference we get
•       So phase difference Ø = ωL – 1/ωC / R
•    = 2Π x 50 x 0.3 – 1 / 2Π x 50 x 15 x 10^-6 / 50
•       = - 118.114 / 50
•        = - 2.36
•         = - 2.4
•         = - 67 degrees
• Since the value is negative we have the current leads the voltage.

Reference link will be

https://brainly.in/question/49339000