- #1
Meow12
- 45
- 20
- Homework Statement
- You have a ##200\ \Omega## resistor, a ##0.400\ H## inductor, a ##6.00\ \mu F## capacitor, and a voltage source that has a voltage amplitude ##30.0\ V## and an angular frequency of 250 rad/s. They are connected to form an L-R-C series circuit.
(a) What are the voltage amplitudes across the resistor, inductor, and capacitor?
(b) Explain how it is possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source.
- Relevant Equations
- ##X_L=\omega L##, ##\displaystyle X_C=\frac{1}{\omega C}##, ##Z=\sqrt{R^2+(X_L-X_C)^2}##
##\displaystyle I=\frac{V}{Z}##
## V_R=IR##, ##V_L=IX_L##, ##V_C=IX_C##
(a) Substituting the values, I get ##X_L=100\ \Omega##, ##X_C=666.67\ \Omega##.
From this, I get ##Z=601\ \Omega##, ##I=49.9\ mA##
##V_R=9.98\ V##, ##V_L=4.99\ V##, ##V_C=33.3\ V##
(b) It's possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source because ##V_R##, ##V_L-V_C##, and ##V## constitute a right triangle where ##V^2=V_R^2+(V_L-V_C)^2##.
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My answer to (a) matches with the one given in the back of the textbook.
But is my answer to (b) correct?
Thanks.
From this, I get ##Z=601\ \Omega##, ##I=49.9\ mA##
##V_R=9.98\ V##, ##V_L=4.99\ V##, ##V_C=33.3\ V##
(b) It's possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source because ##V_R##, ##V_L-V_C##, and ##V## constitute a right triangle where ##V^2=V_R^2+(V_L-V_C)^2##.
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My answer to (a) matches with the one given in the back of the textbook.
But is my answer to (b) correct?
Thanks.