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The expressions for the steady-state voltage and current at the terminals of the circuit seen in the figure are

vg=350cos(5000πt+77∘)V,
ig= 8sin(5000πt+140∘) A
find the impedance seen by the source
b.)by how many microseconds is the current out of phase with the voltage

1 Answer

1 vote

Answer:

To find the impedance seen by the source in this circuit, we can use the equation for impedance, which is given by:

Z = V / I

where Z is the impedance, V is the voltage, and I is the current. In this case, we are given the expressions for the steady-state voltage and current at the terminals of the circuit, so we can substitute those values into the equation to find the impedance.

The voltage expression is given as:

V = 350cos(5000πt + 77°) V

and the current expression is given as:

I = 8sin(5000πt + 140°) A

Substituting these values into the equation for impedance, we get:

Z = (350cos(5000πt + 77°) V) / (8sin(5000πt + 140°) A)

= (350cos(5000πt + 77°) V) / (8sin(5000πt + 140°) A)

= (350cos(5000πt + 77°) V) / (8sin(5000πt + 140°) A)

= 43.75cos(5000πt + 77°) / sin(5000πt + 140°)

Therefore, the impedance seen by the source in this circuit is 43.75cos(5000πt + 77°) / sin(5000πt + 140°).

To find the time delay between the voltage and the current in this circuit, we can compare the phase angles of the voltage and current expressions. The phase angle of the voltage is 77°, and the phase angle of the current is 140°, so the time delay between the voltage and the current is 140° - 77° = 63°.

Since the angular frequency of the voltage and current in this circuit is 5000π radians per second, the time delay between the voltage and current is equal to the phase delay in degrees divided by the angular frequency in radians per second, or 63° / 5000π radians/second = 0.000063 seconds. This time delay is equal to 63 microseconds.

User Chandan Y S
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