An inductor L=1 H is connected across a voltage source with v(t) = 1+1 cos(t) +2 cos(2t) +3.8 sin(t) +1.8 sin(2t).The magnitude of the second harmonic component of the current is:
Question
An inductor L=1 H is connected across a voltage source with v(t) = 1+1 cos(t) +2 cos(2t) +3.8 sin(t) +1.8 sin(2t).The magnitude of the second harmonic component of the current is:
Solution 1
To find the magnitude of the second harmonic component of the current, we first need to understand that the current through an inductor is given by the integral of the voltage across it divided by the inductance.
The voltage v(t) is given as a sum of sinusoidal functions. We are interested in the second harmonic component, which corresponds to the terms with 2t in them.
The second harmonic component of the voltage is 2 cos(2t) + 1.8 sin(2t).
To find the current, we integrate this voltage over time and divide by the inductance.
The integral of cos(2t) is (1/2)sin(2t) and the integral of sin(2t) is -(1/2)cos(2t).
So, the second harmonic component of the current is [(1/2)sin(2t) - (1/2)cos(2t)] / 1 H = 0.5 sin(2t) - 0.5 cos(2t) A.
The magnitude of this current is found using the Pythagorean theorem, sqrt[(0.5)^2 + (-0.5)^2] = 0.5 sqrt(2) A.
So, the magnitude of the second harmonic component of the current is 0.5 sqrt(2) A.
Solution 2
To find the magnitude of the second harmonic component of the current, we first need to understand that the current through an inductor is given by the integral of the voltage across it divided by the inductance.
The voltage v(t) is given as a sum of sinusoidal functions. We are interested in the second harmonic component, which corresponds to the terms with 2t in them.
The second harmonic component of the voltage is therefore 2 cos(2t) + 1.8 sin(2t).
To find the current, we integrate this voltage over time and divide by the inductance.
The integral of cos(2t) is (1/2)sin(2t) and the integral of sin(2t) is -(1/2)cos(2t).
So, the second harmonic component of the current is (1/1)[(1/2)sin(2t) - (1/2)cos(2t)] = 0.5 sin(2t) - 0.5 cos(2t).
The magnitude of a sinusoidal function a sin(t) + b cos(t) is sqrt(a^2 + b^2).
So, the magnitude of the second harmonic component of the current is sqrt((0.5)^2 + (-0.5)^2) = sqrt(0.25 + 0.25) = sqrt(0.5) = 0.707 A.
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