This preview shows pages 1–3. Sign up to view the full content.
Practice Problem 3 Solution in Detail
John McRaven
Figure 1:
3) For the circuit in Figure 1, calculate the equivalent resistance for the
circuit. Calculate the current, voltage, and power for each resistor and the
battery.
Solution:
There are 3 resistors in this circuit. The 4Ω and 12Ω resistors are in
parallel. The 5Ω resistor is in series with the combination of the 4Ω and 12Ω
resistor. So, we must ﬁrst combine the 4Ω and 12Ω resistor using the parallel
resistor equation:
1
R
P
=
1
R
1
+
1
R
2
=
1
12
+
1
4
=
4
12
=
1
3
R
P
= 3Ω
R
P
is now in series with the 5Ω resistor. We use the series equation to
combine them.
R
S
=
R
1
+
R
2
= 5 +
R
P
= 5 + 3 = 8Ω
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document We have now reduced the circuit to one resistor, so 8Ω is the equivalent
resistance.
R
eq
= 5 +
1
1
4
+
1
12
= 8
We can now calculate the total current in the circuit. Whenever the
circuit has been reduced to one battery and one resistor, the current is:
I
=
E
R
eq
=
16
8
= 2
A
This makes the power coming out of the battery:
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 01/05/2011 for the course PHY 7B PHY7B taught by Professor Cebra during the Spring '10 term at UC Davis.
 Spring '10
 Cebra

Click to edit the document details