Download Problem 4.62 - Instructure

Problem 4.62 Part A
Use source transformations to find vo in the circuit.
Express your answer with the appropriate units.
Solution:
Neither the 1A source or the 520V source can be transformed.
All we can do is collapse the circuit by de-activating the current and voltage sources
De-activating the 1A source turns it into an open circuit.
De-activating the 520V source turns it into short circuit.
The equivalent résistance is then: 40 + 4 + 6 = 50 Ohms
The Thevenin voltage is then the open circuit voltage across vo.
By observation, when the resistance across vo is replaced with an open circuit:
There is no current through the 4 ohm or 6 ohm resistor
Therefore, the current through the 260 ohm resistor must be equal and opposite to the current
flowing through the 520V source.
Thus, using KCL at the node directly to the right of the 16 ohm resistor tells us:
Current through the 260 ohm resistor = Current through the 520V source + 1 + V/40
V = -40V
and
Voc = V + 520 = 480V
Thus, using the Thevenin Equivalent circuit to find vo gives:
vo = 480V* (250)/(250 + 50) = 400V
Part B
Find the power developed by the 520V source.
Express your answer with the appropriate units.
The voltage V at the positive terminal of the 520V source, by voltage divider is:
400V = V * 250/(250 + 4 + 6)
V = 416V
which makes the voltage at the minus terminal of the 520V source:
416-520 = -104V
Which makes the current through the 40 ohm resistor:
-104/40 = -2.6A
By KCL, the current I through the 520V source is then:
2.6 + 520/260 = 1A + I
I = 3.6A
And the power developed: 3.6*520 = 1872W
Solution:
Part c
Find the power developed by the 1A source.
Express your answer with the appropriate units.
Solution:
From the previous problem, the voltage across the 1A source is -­‐104V – 1A*16 = -­‐120V Therefore, the power is -­‐120*1 = 120W