PHYS345 Electricity and Electronics

Quiz 4: ac Circuit Analysis

Determine the numerical value of the current through the capacitor by whatever method you prefer.

Circuit of P4.23 modified slightly

Nodal analysis is pleasant enough. Mesh analysis is also not too bad...

Solution

Nodal Analysis

Ground the bottom node.
The nodal equation for the top node is then:
(V1-10)/5 + V1/(-j4) + (V1-5)/(j2) = 0

Multiplying through by j20 yields:
4j(V1-10) - 5V1 + 10(V1-5) = 0
(5+j4)V1 = 50+j40
V1 = 10

Thus the current through the capacitor is:
Ix = 10/(-j4) = j2.5 A

  Mesh Analysis

Mesh 1:
(5-j4)I1 - (-j4)I2 = 10
Mesh 2:
- (-j4)I1 + (-j4+j2)I2 = -5

Simplifying slightly:
1:   (5-j4)I1 + j4I2 = 10
2:   j4I1 - j2I2 = -5

Multiplying the second equation by 2 and
adding to the first equation yields:
(5-j7)I1 = 0!

Thus the current in the first mesh is zero; from equation 2:
I2 = -5/(-j2) = -j2.5 A

The current through the capacitor is I1 - I2, so
Ix = j2.5 A.

Very interesting! There is no current flow through the 10 V source??? It's very happy though...

Thevenin Analysis

If you are a Thevenin enthusiast, then your generator should consist of a voltage source of 5.94 V \/ 16.8 deg and an impedance of 1.86 ohm \/ 68.2 deg. Full speed ahead...

Thanks go to Chris Lane (Fall 1999) for pointing out an important typographical error in the original solution.


"http://www.physics.udel.edu/~watson/phys345/quiz/4soln.html"
Last updated October 27, 1999.
Copyright George Watson, Univ. of Delaware, 1998.