Simple AC Circuits, Reactance and Impedance: Alternating current, as we all know, has become an integral part of our life. Generally, for electricity at our home, we prefer AC supply instead of DC supply due to several reasons. Some of the reasons are as mentioned below:
1. AC can be transmitted over long distances
2. AC is cheaper to generate rather than DC
3. Continuous generation occurs in AC while in DC, you need to replace batteries
4. AC is safe as compared to DC
5. AC can be easily converted to DC while vice versa is difficult
6. AC has a negligible copper loss

There are several other advantages too with only a few disadvantages like it gives an attractive shock while DC gives repelling shock and some storage problem. Else AC is used for maximum electrical devices at home. Nowadays, such devices are developed with works on the hybrid of AC as well as DC. This means that it takes advantage of both AC and DC.

Purely Resistive Circuit

For this circuit, a resistor is connected to an AC supply as shown in the diagram. Here, the voltage supplied to the circuit is given by:
\(V = {V_m}\sin \left( {\omega t} \right)\)
Here, \({V_m}\) is the maximum value of voltage supplied
According to Ohm’s law, we know that current flowing through the resistor will be:
\(i = \frac{V}{R}\)
\(i = \frac{{{V_m}}}{R}\sin \left( {\omega t} \right)\)
\(\therefore i = {i_m}\sin \left( {\omega t} \right)\)
Here, \({i_m} = \frac{{{V_m}}}{R}\) is the maximum value of current passing through the resistor. Now, if you look at the equations of voltage and current, there is no phase difference between them. Hence, the current and voltage in this circuit are said to be in phase with each other. The diagrams shown below are phasor diagrams and parameter vs. time graphs for purely resistive AC circuits.

Purely Inductive Circuit

For this circuit, an inductor is connected to an AC supply as shown in the diagram. Here, the voltage applied by the circuit remains the same as the previous one.
\(V = {V_m}\sin \left( {\omega t} \right)\)
The potential difference across an inductor is given by the equation:
\({V_L} = L\frac{{di}}{{dt}}\)
Hence, the current flowing through this inductor will be:
\(i = \int {\frac{{{V_L}}}{L}dt} \)
\(i = \frac{1}{L}\int {{V_m}\sin \left( {\omega t} \right)dt} \)
\(i = \, – \frac{{{V_m}}}{{\omega L}}\cos \left( {\omega t} \right)\)
\(i = \frac{{{V_m}}}{{{X_L}}}\sin \left( {\omega t – \frac{\pi }{2}} \right)\)
Here, \({X_L} = \omega L\) is known as inductive reactance which is the opposition offered by the inductor to the AC supply.
\(\therefore i = {i_m}\sin \left( {\omega t – \frac{\pi }{2}} \right)\)
Here, \({i_m} = \frac{{{V_m}}}{{{X_L}}}\) is the maximum current which is flowing through the circuit. Now, if you look at the equations of current and voltage across the inductor, you will see that both have a phase difference of \(\frac{\pi }{2}\) between them. Voltage is leading current by an angle of \(\frac{\pi }{2}.\) The phasor diagram and AC curve of the purely inductive circuit are as shown below.

Purely Capacitive Circuit

Here, a capacitor is connected to an AC supply, as shown in the diagram. The voltage supplied to the capacitor is:
\(V = {V_m}\sin \left( {\omega t} \right)\)
The potential difference across the plates of a capacitor is given by the equation:
\({V_c} = \frac{q}{c}\)
\({V_c} = \frac{1}{C}\int {idt} \)
Hence, the current flowing through this capacitor will be:
\(i = C\frac{{dV}}{{dt}}\)
\(i = C\frac{d}{{dt}}\left( {{V_m}\sin \left( {\omega t} \right)} \right)\)
\(i = \omega C{V_m}\cos \left( {\omega t} \right)\)
\(i = \frac{{{V_m}}}{{\frac{1}{{\omega C}}}}\sin \left( {\omega t + \frac{\pi }{2}} \right)\)
\(i = \frac{{{V_m}}}{{{X_c}}}\sin \left( {\omega t + \frac{\pi }{2}} \right)\)
Here, \({X_c} = \frac{1}{{\omega c}}\) is known as the capacitive reactance, which is the oppositive offered by the capacitor to an AC supply.
\(\therefore i = {i_m}\sin \left( {\omega t + \frac{\pi }{2}} \right)\)
Here, \({i_m} = \frac{{{V_m}}}{{{X_c}}}\) is the maximum current that is flowing through the circuit. Now, if you look at the equations of current and voltage through the capacitor, you will see that there is a phase difference of \(\frac{\pi }{2}\) between them. Voltage is lagging behind capacitor by an angle of \(\frac{\pi }{2}.\) The phasor diagram and AC curve for purely capacitive AC circuit is shown below:

RLC Series Circuit

This circuit includes a resistor, an inductor and a capacitor all connected in series with each other as shown in the diagram.
If it was the case of a DC circuit, we would have simply distributed the total voltage supply among resistor inductor and capacitor. But since this is the case of AC supply, things can go a little bit different. Inductor and capacitor are known as the passive elements of the circuit, whereas resistor is known as the active element of the circuit.
Active elements do mean those elements which are able to generate/consume power from the supply, whereas passive elements are those elements that are used to store the power from the supply.

Since all three are connected in series to each other, the current passing through them will be considered to be the same. Now, as we have discussed earlier in this article that voltage and current for resistor are in the same phase; voltage leads current by a phase difference of \(\frac{\pi }{2}\) for an inductor and current leads voltage by a phase difference of \(\frac{\pi }{2}\) for a capacitor, we can draw the phasor diagram for the complete RLC series circuit as below.

Here, in the circuit, we will assume that \({V_L} > {V_C}.\) Hence, looking at the phasor, we can write for the total current supplied to be:
\(V = \sqrt {V_R^2 + {{\left( {{V_L} – {V_V}} \right)}^2}} \)
Substituting, \({V_R} = iR,\,{V_L} = i{X_L}\) and \({V_C} = i{X_C}\) in the above equation
\(V = \sqrt {{{\left( {iR} \right)}^2} + {{\left( {i{X_L} – i{X_C}} \right)}^2}} \)
\(V = i\sqrt {{R^2} + {{\left( {{X_L} – {X_C}} \right)}^2}} \)
Here, \(Z = \sqrt {{R^2} + {{\left( {{X_L} – {X_C}} \right)}^2}} \) is known as impedance of the circuit, which is the overall opposition to the current by all the elements present in the circuit combined.

Since we have assumed that the voltage across the inductor is greater than the voltage across the capacitor, the voltage in the circuit is leading the current by some angle. This phase difference in the circuit can be found using the phasor diagram.
\(\tan \theta = \frac{{{V_L} – {V_C}}}{{{V_R}}}\)
\(\tan \theta = \frac{{i{X_L} – i{X_C}}}{{iR}}\)
\(\tan \theta = \frac{{{X_L} – {X_C}}}{R}\)
Hence, the phasor difference can be given by:
\(\theta = \left( {\frac{{{V_L} – {V_C}}}{{{V_R}}}} \right) = \left( {\frac{{{X_L} – {X_C}}}{R}} \right)\)

If the voltage supply in the circuit is \(V = {V_m}\sin \left( {\omega t} \right),\) then the current supplied to the circuit will be \(i = {i_m}\sin \left( {\omega t – \theta } \right).\) If \({X_L} > {X_C},\) the overall voltage in the circuit is leading current. If \({X_C} > {X_L}.\) the overall current is leading in the circuit.

We can also write the impedance of the circuit as:
\(Z = \sqrt {{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}} \)
From this equation, we can say that the impedance of the circuit can be controlled by controlling the frequency in the circuit. And by controlling impedance, we can even control the current that is supplied to the circuit.

Summary

From this article, we came to know about the advantages and disadvantages of AC over DC. Then we gained some basic knowledge about the working of an AC circuit by connecting a resistor, capacitor and inductor individually to the supply. In the case of a resistor, the voltage and current are together in the same phase to each other. This element is considered to be the active element of the circuit. Whereas in the case of capacitor and inductor, the phase difference between current and voltage is \(\frac{\pi }{2}\) where either the current is leading, or the voltage is leading. Here, we came to know about the inductive reactance and capacitive reactance, which is the opposition offered to the current by inductor and capacitor, respectively, in an AC circuit. Inductor and capacitor are considered to be the passive elements of an AC circuit.

Finally, we combined all the three elements of the circuit together in series with each other and studied the RLC series circuit, where we learned about impedance which shows the overall opposition offered to the current supply by all the elements combined together and looked into the equation to find phase difference.
Impedance for RLC series circuit: \(Z = \sqrt {{R^2} + {{\left( {{X_L} – {X_C}} \right)}^2}} \)
Phase difference between current and voltage for RLC series circuit:
\(\theta = \left( {\frac{{{V_L} – {V_C}}}{{{V_R}}}} \right) = \left( {\frac{{{X_L} – {X_C}}}{R}} \right)\)

Frequently Asked Questions

Q.1. What is reactance?
Ans: Reactance is the opposition offered to the current by an element in an AC circuit. If it is the case of a DC supply, we only need to consider resistance, but for AC supply, we need to consider active as well as passive elements.

Q.2. What is \({{X_L}}\) and \({{X_C}}\)?
Ans: \({X_L} = \omega L\) is the inductive reactance offered by the inductor while \({X_C} = \frac{1}{{\omega C}}\) is the capacitive reactance offered by the capacitor.

Q.3. How do you calculate impedance from reactance?
Ans: The formula to calculate impedance using reactance is \(Z = \sqrt {{R^2} + {{\left( {{X_L} – {X_C}} \right)}^2}} .\)

Q.4. What do you mean by impedance?
Ans: Impedance is the opposition offered to the electrical flow in an AC circuit by all the elements combined.