Question Video: Resonance in Alternating Current Circuits | Nagwa Question Video: Resonance in Alternating Current Circuits | Nagwa

# Question Video: Resonance in Alternating Current Circuits Physics • Third Year of Secondary School

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What is the resonant frequency of the circuit shown in the diagram?

02:49

### Video Transcript

What is the resonant frequency of the circuit shown in the diagram?

The circuit consists of an alternating-voltage source connected to a series combination of a 35-Ξ© resistor, a 7.5-henry inductor, and a 350-microfarad capacitor. And weβre asked to find the resonant frequency of this circuit. Recall that the inductive reactance in a circuit is the angular frequency of the voltage source times the inductance. And the capacitive reactance is one divided by the angular frequency of the voltage source times the capacitance. On resonance, these two reactances are equal.

If we call the resonant angular frequency π naught, then we have that π naught πΏ is equal to one over π naught πΆ, which we can solve for π naught. When we solve this equation for π naught, we find that the resonant angular frequency is equal to one divided by the square root of the inductance of the inductor times the capacitance of the capacitor. Now, this is a formula for angular frequency, but weβre looking for just regular frequency. So we need to use the relationship that angular frequency is two π times the regular frequency.

Alright, so letβs plug our definition for angular frequency into our equation for the resonant angular frequency. We have two times π times the resonant frequency is equal to one divided by the square root of the inductance times the capacitance. To get this expression into the final form we need, we simply divide both sides by two π. On the left-hand side, two π divided by two π is one, and weβre just left with π naught. On the right-hand side, the two π just becomes part of the denominator of our fraction. This leaves us with the final formula we need. Resonant frequency is equal to one divided by two π times the square root of the inductance, in henries, times the capacitance, in farads.

So now we just need to plug in values. We have an inductance in henries. Itβs 7.5 henries. However, our capacitance is given in microfarads instead of farads. To convert to farads, recall that there are one million microfarads per farad. In other words, one microfarad is equivalent to 10 to the negative sixth farads. Since we have 350 microfarads, our capacitance is equivalent to 350 times 10 to the negative sixth farads.

Plugging our inductance and capacitance into our formula for resonant frequency, this gives us one divided by two π times the square root of 350 times 10 to the negative sixth farads times 7.5 henries. It turns out that the square root of one farad times one henry is one second. So we can rewrite the denominator with units of seconds.

Now, one divided by seconds is the unit hertz, which is used for frequency. So now we have an expression for the resonant frequency. Thatβs a number times a unit hertz, which is the right unit for frequency. So now all we have to do is evaluate this number with a calculator. When we do this evaluation, we find that the entire numerical expression is approximately equal to 3.1. So the resonant frequency of this circuit is 3.1 hertz. Itβs worth noting that the 35-Ξ© resistor played no role in our calculation of the resonant frequency.

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