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The circuit in this problem has two resistors, one capacitor, and one inductor.
The power supply is a sinusoidal voltage source with an amplitude of 4 volts and a
frequency of 3,000 radians per second.
We want to find the apparent power absorbed by the load in the circuit.
If we can find the voltage across the load, as well as the current that flows through the
load, it is possible to calculate the average power, P, and the reactive power, Q.
The complex power is P plus jQ.
The apparent power is defined as the magnitude of the complex power.
An easier way to find apparent power is to find the product of the RMS values of the
voltage and current.
So the complex power is the RMS value voltage multiplied by the RMS value of the
So here we need to get the voltage and current for the load in the circuit.
Let’s construct the phasor-domain equivalent circuit.
The phasor transform for the voltage should be represented by this equation.
For the resistors, the impedance is just the resistances.
We find the capacitance using this equation.
[math equation]
We find the inductance using this equation.
[math equation]
Now we need to find the load voltage.
In the phasor domain, the voltage across the node is called Vab.
It is possible to combine the impedances of the two parallel branches.
Zab is the impedance of the capacitor in parallel with the resistor and inductor.
[math equation]
That is the product of the two impedances divided by the sum of the two impedances.
[math equation]
And here is our result.
The current through Zab should be the voltage divided by the total impedance in the
The voltage across the load should be the current multiplied by Zab.
[math equation]
So this is our voltage for Vab.
We know the voltage across a and b; we can find the load current easily using this
[math equation]
And here is our load current.
The unit is milliamps.
Because voltage is in volts, resistance here is in kilohms.
So we know the load voltage and current.
Apparent power is the RMS value for the load voltage multiplied by the RMS value of
the current.
We know the amplitude for V, so we can divide by the square root of 2 to get the RMS
We do the same for I.
So the apparent power
is 0.75 millivolt-amps.