A six-ohm resistor, a capacitor with reactance of 80 ohms, and an inductor with
negligible resistance and a self-inductance coefficient of 0.28 henries are
connected in series with an AC power supply of 20 volts at 50 hertz. Calculate the maximum current in the circuit, treating 𝜋 as exactly 3.14.
Let’s start out by sketching this AC circuit. The elements in this electrical circuit are an AC power supply, a resistor, a
capacitor, and an inductor. We’re given some information about each of these elements and we want to calculate
the maximum current that ever runs through this circuit.
To do that, let’s consider a relationship between current, voltage, and the overall
resistance of this circuit. The maximum current ever running through a circuit like this is equal to the maximum
voltage supplied by the power source divided by the impedance 𝑍 of the circuit. This impedance is analogous to resistance. And it comes up and we consider AC-power-supplied circuits.
In general, impedance has three components that can contribute to it. The first is the resistance of the circuit, the second is the inductive reactance 𝑋
sub 𝐿, and the third is the capacitive reactance 𝑋 sub 𝐶 of the circuit.
Based on our equation for 𝐼 sub max, which is what we want to solve for, we know
that to do that, we’ll have to solve for the maximum voltage supplied as well as the
impedance in the circuit. Let’s start with the maximum voltage 𝑉 sub max.
In our problem statement, we’re given the voltage of our power supply at 20
volts. However, this is not equal to 𝑉 sub max. When the voltage supplied to a circuit varies as it does in this case with an AC
power supply, it’s common to give the voltage of that power supply at a level that’s
known as the effective voltage 𝑉 sub eff. This effective voltage is neither the highest nor the lowest voltage supplied by the
sinusoidal power supply.
So 𝑉 sub eff and 𝑉 sub max are not the same thing. But thankfully, there is an equation that does relate these two. If we take our given value for the effective voltage and multiply it by the square
root of two over two, then that value will be equal to the maximum voltage output by
the power supply.
If we plug in 20 volts for the effective voltage and calculate, we find that the
maximum voltage is equal to 28.2885 volts. That’s the most voltage that our power supply will ever create.
Now that we know the numerator of the fraction which will tell us the maximum
current, we’ll look to solve for the denominator, the impedance of our circuit. In our problem statement, we’re told that the resistor in this circuit has a
resistance of six ohms and that the capacitor has a reactance of 80 ohms.
In other words, we’ve just been told capital 𝑅 and 𝑋 sub 𝐶 for this circuit. However, we’re never told 𝑋 sub 𝐿, the inductive reactance. To solve for that term, we’ll need to use the fact that 𝑋 sub 𝐿 is equal to two
times 𝜋 times the frequency of oscillation of the circuit multiplied by the
self-inductance coefficient capital 𝐿.
The frequency of the circuit 𝑓 is given as 50 hertz and capital 𝐿 is given as 0.28
henries. Treating 𝜋 as exactly 3.14, when we multiply all these numbers together. We find that 𝑋 sub 𝐿, the inductive reactance, is 87.92 ohms.
Now that we know that value, we have all the components that go into making up the
impedance of the circuit 𝑍. When we plug in for these three values, we use six ohms for our resistance 𝑅 87.92
ohms for inductive reactance and 80 ohms given in our problem statement as the
capacitive reactance. Combining these terms this way gives us an overall circuit impedance of 9.936
What we’ve now accomplished is we’ve solved for the numerator and denominator that
we’ll need in order to calculate the maximum current in the circuit. That maximum current is equal to 28.2885 volts divided by 9.936 ohms. That value is equal to 2.847 amps. That’s the maximum current in this circuit.