ECE 53a
Quiz #1 SOLUTIONS
Pamela Cosman
1/29/08
Problem 1: (This is problem 1 from homework 2.)
V
a
is just the same as the source voltage. It
is 8V. And for
V
o
, we see it is just a voltage divider. We have to notice that
V
o
has the opposite
polarity from
V
a
= 8
V
, so we need to include a minus sign in the voltage division equation:
V
o
=

2
2 + 3 + 3
×
8 =

2
V
Problem 2: (This is similar to problem 3 in the practice quiz.)
+
_
+

2V
I
x
x
0.5 I V
1
1.5
First to ±nd the current, we note that the total voltage powering the circuit is the sum of the two
voltage sources in series: 2 + 0
.
5
I
x
. The total resistance in the circuit is the sum of the two series
resistors: 1 + 1
.
5 = 2
.
5.
By Ohm’s law, the current that ²ows will be the ratio:
I
x
=
2 + 0
.
5
I
x
2
.
5
and we solve this equation to obtain:
I
x
= 1
A
The fastest way to answer this for the resistors is to remember that one expression for power
absorbed by a resistor is
p
=
Ri
2
Since
I
x
= 1, and it is the same current throughout the circuit, the power absorbed by the 1 Ohm
resistor is
1
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 Spring '08
 Cheng
 Volt, Resistor, Voltage source, rth

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