11. AC Circuit Exercises
by M. Bourne
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.
Note the units in the question: `"mA"` and `"k"Omega`.
Applying the formula for voltage:
We can move the powers of 10 outside the absolute value (magnitude) signs.
The powers of 10 cancel out.
We now proceed to find the magnitude of each complex number. (See Impedance & Phase Angle for background.)
So the voltage is `38\ "V"`.
Get the Daily Math Tweet!
IntMath on Twitter
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?
The total reactance across a circuit with a resistance, an inductance and a capacitance in series is given by:
VRLC = VR + j(VL − VC) (from the definitions)
`|V_R| = 12\ "V"`
`|V_L| = 15.5\ "V"`
`|V_C| = 10.5\ "V"`
VRLC = VR + j(VL − VC)
= 12 + j(15.5 − 10.5)
= 12 + 5j V
The magnitude of this voltage is:
|12 + 5j| = 13 V (from calculator)
Easy to understand math videos: