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11. AC Circuit Exercises

by M. Bourne

The following exercises make use of what you learned in Definitions and Impedance and Phase Angle, as well as the Complex Number Basic Operations and Products and Quotients sections.

Exercise 1

Given that the current in a given circuit is 3.90 - 6.04j mA and the impedance is 5.16 + 1.14j kΩ, find the magnitude of the voltage.

Answer

Note the units in the question: `"mA"` and `"k"Omega`.

Applying the formula for voltage:

`|V|=|I|xx|Z|`

`=|(3.90-6.04j)xx10^-3|` `xx|(5.16+1.14j)xx10^3|`

We can move the powers of 10 outside the absolute value (magnitude) signs.

`=|(3.90-6.04j)|xx10^-3` `xx|(5.16+1.14j)|xx10^3`

The powers of 10 cancel out.

`=|3.90-6.04j|xx|5.16+1.14j|`

We now proceed to find the magnitude of each complex number. (See Impedance & Phase Angle for background.)

`=sqrt((3.90)^2+(-6.04)^2)` `xxsqrt((5.16)^2+(1.14)^2)`

`=7.190xx5.284`

`=38.0\ text(volts)`

So the voltage is `38\ "V"`.

Exercise 2

A resistor, an inductor and a capacitor are connected in series across an ac voltage source. A voltmeter measures `12.0\ "V"`, `15.5\ "V"` and `10.5\ "V"` respectively, when placed across each element separately. What is the magnitude of the voltage of the source?

Answer

The total reactance across a circuit with a resistance, an inductance and a capacitance in series is given by:

VRLC = VR + j(VLVC) (from the definitions)

Now

`|V_R| = 12\ "V"`

`|V_L| = 15.5\ "V"`

`|V_C| = 10.5\ "V"`

Now

VRLC = VR + j(VLVC)

= 12 + j(15.5 − 10.5)

= 12 + 5j V

The magnitude of this voltage is:

|12 + 5j| = 13 V (from calculator)