Video: Resistive-Inductive-Capacitive Alternating Current Circuit Characteristics

The figure shows an RLC circuit. What is the total impedance of the circuit? What is the phase angle between the current and emf in the circuit?

04:45

Video Transcript

The figure shows an RLC circuit. What is the total impedance of the circuit? What is the phase angle between the current and emf in the circuit?

In part one, we want to solve for the circuit impedance. We’ll call that capital 𝑍. And in part two, we want to solve for the phase angle that exists between the current and emf in the circuit. We’ll call that angle πœƒ. As we look at our circuit, we see it has a resistor and inductor and a capacitor. And it also has an AC, or alternating current, power supply.

We can begin solving for the impedance 𝑍 by recalling the definition for impedance. Impedance is equal to circuit resistance plus circuit reactance. In general, it’s the total sum of all forces that oppose the flow of current in a circuit. When we talk about reactance, we want to know the reactance of both inductors and capacitors to match with our circuit components. For a capacitor, the reactance is equal to one over two πœ‹ times the frequency of oscillation of the circuit 𝑓 times the capacitance 𝐢. And for an inductor, the reactance is equal to two πœ‹ 𝑓 times 𝐿.

Going back to our equation for impedance 𝑍, for an RLC circuit like the one we have here, 𝑍 is equal to the square root of the resistance squared plus the reactance of the inductor minus the reactance of the capacitor quantity squared. This means that we can solve for the impedance if we know the resistance 𝑅, the capacitance 𝐢, the inductance 𝐿, and the frequency 𝑓 of our circuit. Based on the circuit diagram, we already know the resistance, inductance, and capacitance. But we still need to solve for the frequency 𝑓.

To do that, let’s look at our expression for the AC power supply. And in particular, we’ll look at the factor that multiplies 𝑑 in our sine function. This factor, 120πœ‹, is equal to the angular frequency of the power supply. When we recall that 𝑓 is equal to πœ” over two πœ‹, that means that in our case, 𝑓 is equal to 120πœ‹πœ” over two πœ‹ hertz or 60 hertz. That’s the oscillation frequency of this power supply.

We’re now ready to solve for the reactance of the inductor, the reactance of the capacitor, and ultimately for the impedance 𝑍. When we plug in all these values to solve for impedance, using 5.0 ohms for 𝑅, 60 hertz for the frequency 𝑓, 25 times 10 to the negative third henries for 𝐿, and 400 times 10 to the negative sixth farads for 𝐢.

When we enter all these values on our calculator, we find that, to two significant figures, 𝑍 is 5.7 ohms. That’s the impedance of this RLC circuit.

Now we move on to solving for πœƒ which is the phase angle between the current and the emf in our circuit. If we were to create a vector map of our impedance equation, then with impedance 𝑍 as the hypotenuse of a right triangle, 𝑅 the resistance would be the lower leg. And the difference in inductive and capacitive reactance would be the vertical leg. In this geometry, the angle formed in the lower left-hand corner is πœƒ. This is the angle that represents the phase difference between current and emf in the circuit.

Looking at this triangle, we can see that the tangent of the angle πœƒ is equal to the vertical leg, π‘₯ sub 𝐿 minus π‘₯ sub 𝐢, divided by the horizontal leg, the resistance 𝑅. If we take the inverse tangent of both sides of this equation, we see that πœƒ is equal to the arc tangent of π‘₯ sub 𝐿 minus π‘₯ sub 𝐢 over the resistance 𝑅. We can expand this expression in parentheses based on our understanding of the equations for π‘₯ sub 𝐢 and π‘₯ sub 𝐿.

When we plug in for the values of frequency, inductance, capacitance, and resistance to this equation and evaluate the inverse tangent of the expression, we find that, to two significant figures, it equals 29 degrees. This is the phase angle relationship between current and emf.

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