# RLC circuit. Calculate RMS voltage drop across R, L, and C?

Consider a series RLC circuit with R = 300 ohm, L = 30 mH (.03H), and C = 5uF (5*10^-6F). A 3 V peak to peak signal is placed across the circuit with a frequency of 100 Hz.

a) what is the rms voltage drop across the resistor?

b) what is the rms voltage drop across the inductor?

c) what is the rms voltage drop across the capacitor?

d) what is the total power used by the entire circuit?

e) what frequency should be used for maximum power? what is this power?

I'm lost to be honest. any help is appreciated.

Relevance

Given:

R = 300 ohm

L = 0.03 H

C = 5*10^-6 F

Vrms = 3V

f = 100 Hz

In order to find the voltages (V = IR, V = iXL, and V = IXC), you'll need to find the current (I). And before you find the current, you'll need to find the impudence (Z). And to find the impudence, you'll need to find inductive reactance (XL) and the capacitive reactance (XC)... sucks right?

XL = 2 * pi * f * L = 2 * pi * 100 Hz * 0.03 H = 18.8 ohms

XC = 1 / (2 * pi * f * C) = 1 / (2 * pi * 100 Hz * 5.6*10-6) = 318 ohms

Z = sqroot (R^2 + (XL - XC)^2) = sqroot (300^2 + (-299)^2) = 423.7 ohms

Vrms = Irms Z therefore I = 3V/423.7ohms = 0.0071 A

This is a long problem, so I'll let you do the rest:

a.) V = I R

b.) V = I XL

c.) V = I XC

d.) Prms = I^2 R

e.) Not sure for this one, but I think if XL = XC, then you get maximum power.

So, find f when XL = XC, then solve everything again using this new value for the maximum power.

Good luck.

• joshua4 years agoReport

3V is pk-pk not rms. Gotta divide by root2 right?

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• Rlc Circuit Formulas

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• hello

any one can help to find the solution of this question

For the given series RLC Circuit, find voltage drop across resistors, capacitors and inductor.

R1= 5ohm

L1=2Ω

C1=100 µF

V=10 cos (500 t)

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