**Examples for AC circuits**

Example #1

Problem:

An AC circuit carries an rms current of 7.0 Amps. The current travels through a 12 Ohm resistor.

a.) What is the peak current?

Solution: Multiply the rms current by sqrt(2).

* I*
= 9.90

b.) What is the power dissipated in the resistor?

Solution: You can use the normal formula, * i^{2}R*, provided that you use
the rms value of the current.

588 *W*

c.) What is the peak voltage drop across the resistor?

Solution: Use Ohm's law, being careful to use the peak current to get the peak voltage.

118.8 *V*

Example #2

Problem:

Consider the LC circuit to the right. If
one needs to tune this circuit to a frequency of 84 *kHz*, and the capacitor has a
capacitance * C* = 3.0 m

Solution:

Use the relation, .
Solve for * L*.

* L*
= 1.20E-6

Example #3

Problem:

a.) What is the impedance of the circuit to
the right if: * f* = 60

Solution:

* Z*
= 8.54

b.) If the r.m.s. voltage of the source is * V_{rms}*
= 110

Solution:* I_{rms}* =

* I_{rms}* = 12.9 amps

c.) What is the peak current?

Solution:* I_{p}* = sqrt(2) ·

* I_{p}* = 18.2 amps

d.) What is the power dissipated in the resistor?

Solution: P = * I^{2}_{rms}*·

* P*
= 1320

Example #4

Problem:

a.) What is the capacitance such that the current through the circuit is a maximum?

**DATA: f** =
60

Solution: Choose the capacitance such that the reactance of the capacitor cancels the reactance of the inductor:

* C* = 8.8E-5

b.) What
is the r.m.s. current through the circuit for the capacitance found in part *a*?

Solution: The impedance of of the circuit is determined completely by the resistor at
resonance, * Z = R*. The current is then

* I*
= 27.5

c.) Find the capacitance to make the impedance equal to 8 Ohms.

Solution: The impedance is:

First solve for the reactance of the capacitor (** X_{c}**
= .612

* C*
= 4.33E-3

*
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