What Causes Voltage Drops Across an Inductor in an R-L-C Series Circuit?

[Moore65]

Trouble with the R-L-C series. Please Help!

I'm having trouble trying to understand this problem.

The R-L-C series circuit is driven with a AC source with EMF=Vo*sin(wt) where Vo=110.0V and f=60.0 Hz. If R = 20.0 Ohms, L=5.0 * 10^-2 H, and C=50.0uF, find the potential drops across the inductor at to=0 and at a time t1=the first time after to that the EMF reaches a maximum.

The current moves throught the circuit first through the resistor, then through the capacitor, and finally through the "L" before completely the cicuit.

I'm not really sure what to do, can anyone help?

 

[wais]

Hi there,

I can definitely try to help you with this problem. First, let's break down the problem into smaller steps.

Step 1: Find the impedance of the circuit
We know that the impedance of a series circuit is given by the formula Z = √(R^2 + (XL - XC)^2). In this case, XL is the inductive reactance and XC is the capacitive reactance. We can calculate these values using the formulas XL = ωL and XC = 1/ωC, where ω = 2πf. Plugging in the given values, we get XL = 6Ω and XC = 63.7Ω. Now, we can calculate the impedance as Z = √(20^2 + (6 - 63.7)^2) = 65Ω.

Step 2: Calculate the current
Using Ohm's law, we can calculate the current in the circuit as I = V/Z, where V is the source voltage. Plugging in the given values, we get I = 110/65 = 1.69A.

Step 3: Find the potential drop across the inductor at t = 0
At t = 0, the current is just starting to flow through the circuit. This means that the voltage drop across the inductor is equal to the voltage drop across the resistor, since there is no voltage drop across the capacitor yet. Using Ohm's law, we can calculate the voltage drop across the resistor as VR = IR = 1.69 * 20 = 33.8V. This means that the voltage drop across the inductor at t = 0 is also 33.8V.

Step 4: Find the potential drop across the inductor at t = t1
At t = t1, the EMF reaches its maximum, which means that the voltage drop across the capacitor is also at its maximum. This means that the voltage drop across the inductor is now equal to the voltage drop across the resistor and the capacitor in series. Using the voltage divider rule, we can calculate the voltage drop across the inductor as VL = VR * XC/(XL + XC) = 33.8 * 63.7/(6 + 63.7) = 32.7V.

I hope this helps you understand the problem better and find the potential drops across the in