An inductor 20mH, a capacitor 50microF and a resistor of 40ohm are connected In series across a source of emf V=10sin340t. What is the power loss in AC circuit ?

Question

Answer

Calculation of Power Loss in AC Circuit

Given Data:

Inductor (L) = 20mH

Capacitor (C) = 50microF

Resistor (R) = 40ohm

EMF (V) = 10sin340t

Explanation:

When an AC circuit is connected in a series, the current flowing through different components is in phase, and the voltage across each component is not the same. The voltage across the capacitor leads the current by 90 degrees, whereas the voltage across the inductor lags the current by 90 degrees. The voltage across the resistor is in phase with the current.

Using Ohm's law, we can find the current flowing through the circuit:

I = V/Z

Where Z is the impedance of the circuit, given by:

Z = R + j(XL - XC)

Here, XL is the inductive reactance given by XL = 2πfL, and XC is the capacitive reactance given by XC = 1/2πfC.

Using these equations, we can find the impedance Z and the current I flowing through the circuit. The power loss in the circuit can be calculated using the formula:

P = I^2R

Calculation:

Using the given values, we can calculate the impedance of the circuit:

XL = 2πfL = 2π(340)(20x10^-3) = 4.28ohm

XC = 1/2πfC = 1/(2π(340)(50x10^-6)) = 93.46ohm

Z = R + j(XL - XC) = 40 + j(-89.18)ohm

Using Ohm's law, we can find the current flowing through the circuit:

I = V/Z = 10sin340t/(40 - j89.18)A

The power loss in the circuit can be calculated using the formula:

P = I^2R = (10sin340t/(40 - j89.18))^2 x 40W

Answer:

The power loss in the AC circuit is given by the expression (10sin340t/(40 - j89.18))^2 x 40W.

Answer

Ans:- 0.5W..

Answer

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