# RLC series circuit (resonance)

#### jonnym92

Joined Sep 8, 2015
6
A 40uF capacitor in series with coil of resisatance 8 ohms and inductance 80mH is conne cted to 200V 100Hz supply
Calculate
a) circuit impedance- answer z=8+j10.48 xL=50.27 xc=39.79
b) current flowing- answer- 15.17 with angle -52.64
c)voltage across coil answer 762.6 with angle 37.36
d) volts across capacitor answer 603.61 with angle -142.64

those answers are correct according to my tutor the next question is where i am stuck

What additional impedance would need to be added to the circuit for resonance to occur?

now i know XC must equal XL and that there is a formula for resonant frequancy and q-factor etc however i am usure of this extra impedance it is asking for. the questions are then thge same as above but to find out the new values.

any help to find the impedance needed would be a great help as i only attend college one a week as part of work so i am unable to seek help from teachers once at home and this is referall work.

#### Papabravo

Joined Feb 24, 2006
15,041
At the frequency of interest you need to either decrease XL from 50.27 down to 39.79, or increase XC from 39.79 to 50.27, or adjust both to some new value where they are equal at the frequency of interest. At least that is the way I see it. I could be wrong however, since it has happened before.

#### jonnym92

Joined Sep 8, 2015
6
To do this do you just reduce the value of say xl then rearrange the equation to get the new value of L using the values stated in the question?

#### MrAl

Joined Jun 17, 2014
7,958
Hi,

The resonant frequency is:
w=1/sqrt(L*C)

and because they are asking you to add an impedance that means you have to figure out which type of element to add in series in order for w to reach 2*pi*100.
Since two inductors in series result in a total larger inductance and two caps in series result in a total smaller capacitance, you have to figure out if you need a larger inductance or smaller capacitance.
It should be obvious from the above formula that if you add another L the denominator gets larger so the frequency goes down, and if you reduce C (with another cap in series) then the frequency goes up.
This means that if the resonant frequency now is higher than needed, you need to add inductance which means another inductor in series. If the resonant frequency now is lower than needed, then you need to add capacitance in series so the resonant frequency goes up.
So it's not that hard to do you just have to do a calculation to find out which type of element to add in series, then calculate the new value. You can then calculate the value to be added by using the fact that inductors add in series:
L=L1+L2

and the total capacitance goes down with an added cap in series:
C=C1*C2/(C1+C2)

Using whichever formula you need you can calculate C2 knowing C1 already, or L2 knowing L1 already in the circuit.

#### jonnym92

Joined Sep 8, 2015
6
Hi,

The resonant frequency is:
w=1/sqrt(L*C)

and because they are asking you to add an impedance that means you have to figure out which type of element to add in series in order for w to reach 2*pi*100.
Since two inductors in series result in a total larger inductance and two caps in series result in a total smaller capacitance, you have to figure out if you need a larger inductance or smaller capacitance.
It should be obvious from the above formula that if you add another L the denominator gets larger so the frequency goes down, and if you reduce C (with another cap in series) then the frequency goes up.
This means that if the resonant frequency now is higher than needed, you need to add inductance which means another inductor in series. If the resonant frequency now is lower than needed, then you need to add capacitance in series so the resonant frequency goes up.
So it's not that hard to do you just have to do a calculation to find out which type of element to add in series, then calculate the new value. You can then calculate the value to be added by using the fact that inductors add in series:
L=L1+L2

and the total capacitance goes down with an added cap in series:
C=C1*C2/(C1+C2)

Using whichever formula you need you can calculate C2 knowing C1 already, or L2 knowing L1 already in the circuit.
I kind of understand what your saying but still don't quite know what to do! I don't get how you chose the right values to get the 2*pi*100 value
Thanks for the help

#### jonnym92

Joined Sep 8, 2015
6
Calculated that resonance frequency is 559 rounded. I still don't know how to make xl and xc equal each other as surley changing some values will change the resonance frequency? Wish I'd listened in class

#### Papabravo

Joined Feb 24, 2006
15,041
Well here is the thing. The answer may not be unique. All that is required is to figure out an L and a C that will be resonant at the frequency of interest. Then figure out how to get there from where you are.

#### RBR1317

Joined Nov 13, 2010
603
What additional impedance would need to be added to the circuit for resonance to occur?
You may be over-thinking the problem. You have calculated xL=50.27 & xC=39.79 and recognized that at resonance xL=xC.

So what additional impedance would need to be added to xL or xC to make xL=xC at a frequency of 100Hz?

When you figure out how many ohms of reactance need to be added, does that answer the question?

#### MrAl

Joined Jun 17, 2014
7,958
Calculated that resonance frequency is 559 rounded. I still don't know how to make xl and xc equal each other as surley changing some values will change the resonance frequency? Wish I'd listened in class
Hello again,

You have 80mH and 40uF so the angular resonant frequency w is 559 but you have to divide that by 2pi to get the frequency F in Hertz. Dividing like that we get 88.97Hz or about 89Hz.

Now we see that 89 is less than 100, so we need to increase the resonant frequency. There are two ways to do this but we must ADD an impedance in series so we look at the formula for resonance:
w=1/sqrt(L*C)
which means the frequency F is:
F=1/(2*pi*sqrt(L*C))

We see the square root of both L and C in the denominator, and when the denominator increases the left hand side (F) goes down, so we know right away we must decrease the denominator.
Breaking that formula up we have:
F=1/(2*pi*sqrt(L)*sqrt(C))

and we know that connecting two inductors in series means the total inductance goes up, so if we add an inductor the denominator will go up and thus the frequency (left side) will go down, and that will not help. We also know that when we connect two caps in series the total capacitance goes down as per:
C=C1*C2/(C1+C2)

and we know that C1 is 40uf, so all we have to do is find what C2 should be.

First, solving the resonant formula for C we have:
F=1/(2*pi*sqrt(L)*sqrt(C))
sqrt(C)=1/(2*pi*F*sqrt(L))
squaring both sides we get:
C=1/(4*pi^2*F^2*L)

Plugging in the values for the desired new F and the existing L we get:

Thus the total capacitance needed (C) is 31.66uF. Right now the cap value is 40uF, so we must reduce the total capacitance in order to reach the resonant frequency desired.

Now we have to calculate what C2 should be so we use the capacitance in series formula:
C=C1*C2/(C1+C2)

and solve for C2 we get:
C2=(C*C1)/(C1-C)

Now you just plug in the values for C1 (already in the circuit) and the value for C we calculated above. I assume you know how to do this and you should at least try doing it. This new cap C2 is then placed in series with L and C that are already in the circuit, which effectively reduces the original value of C.

So to summarize, we first found out what component needed to be changed (L or C) and then we determined how much it had to change, then calculated the new value that needed to be added to the circuit in series in order to get to the right resonant frequency.
Note we did not have to really deal with making xL=xC, but we could have done it that way too. We could then use trial and error to see what component (L or C) caused the frequency to go up a little so we could reach 100Hz.

If you still have questions feel free to ask right here

#### jonnym92

Joined Sep 8, 2015
6
Hello again,

You have 80mH and 40uF so the angular resonant frequency w is 559 but you have to divide that by 2pi to get the frequency F in Hertz. Dividing like that we get 88.97Hz or about 89Hz.

Now we see that 89 is less than 100, so we need to increase the resonant frequency. There are two ways to do this but we must ADD an impedance in series so we look at the formula for resonance:
w=1/sqrt(L*C)
which means the frequency F is:
F=1/(2*pi*sqrt(L*C))

We see the square root of both L and C in the denominator, and when the denominator increases the left hand side (F) goes down, so we know right away we must decrease the denominator.
Breaking that formula up we have:
F=1/(2*pi*sqrt(L)*sqrt(C))

and we know that connecting two inductors in series means the total inductance goes up, so if we add an inductor the denominator will go up and thus the frequency (left side) will go down, and that will not help. We also know that when we connect two caps in series the total capacitance goes down as per:
C=C1*C2/(C1+C2)

and we know that C1 is 40uf, so all we have to do is find what C2 should be.

First, solving the resonant formula for C we have:
F=1/(2*pi*sqrt(L)*sqrt(C))
sqrt(C)=1/(2*pi*F*sqrt(L))
squaring both sides we get:
C=1/(4*pi^2*F^2*L)

Plugging in the values for the desired new F and the existing L we get:

Thus the total capacitance needed (C) is 31.66uF. Right now the cap value is 40uF, so we must reduce the total capacitance in order to reach the resonant frequency desired.

Now we have to calculate what C2 should be so we use the capacitance in series formula:
C=C1*C2/(C1+C2)

and solve for C2 we get:
C2=(C*C1)/(C1-C)

Now you just plug in the values for C1 (already in the circuit) and the value for C we calculated above. I assume you know how to do this and you should at least try doing it. This new cap C2 is then placed in series with L and C that are already in the circuit, which effectively reduces the original value of C.

So to summarize, we first found out what component needed to be changed (L or C) and then we determined how much it had to change, then calculated the new value that needed to be added to the circuit in series in order to get to the right resonant frequency.
Note we did not have to really deal with making xL=xC, but we could have done it that way too. We could then use trial and error to see what component (L or C) caused the frequency to go up a little so we could reach 100Hz.

If you still have questions feel free to ask right here
Thanks for that it's helped a lot. After doing more work last night I got that it had to be a cap and I think it worked out around 151uf if I remember correctly. Wish I'd found this sight years ago, learned more here in 2 days that 5 year at college haha

#### Russmax

Joined Sep 3, 2015
82
jonnym,

Your problem says you have to ADD a single component to achieve resonance.
You chose to put a cap in series with your other cap.
151 uF is correct (actually 152 uF--careful with rounding).

#### MrAl

Joined Jun 17, 2014
7,958
Hello again,

The way the question was worded it sounded like they wanted an impedance added to the entire circuit, not just one element of the circuit, but if the latter is true then the other alternative is to add an inductor in parallel to the original inductor as that will reduce the total inductance in the resonant formula:
w=1/(sqrt(L)*sqrt(C))

and again as the denominator on the right gets lower the left side becomes larger (what we needed to get to 100Hz) and thus we could calculate a parallel inductor that would also solve the problem.
So instead of using the formula for two caps in series we would use the formula for two inductors in parallel, but then the second inductor MUST be put in parallel with the inductor alone, not in parallel with the circuit. Using the cap solution we can put the extra cap in series with anything including the entire circuit.

I ill leave this as an exercise for the OP

#### Russmax

Joined Sep 3, 2015
82
Which is why you can't choose the parallel inductor method. The coil R and L are distributed. There's no such thing as "in parallel with the inductor alone."