Physics 261 Class Exercise: Circuits - KEY
1. Joe the junior electrician has only a limited supply of electrical components at his disposal. He has two 6V batteries, a large box of 4 ohm resistors, and a large box of 10 ohm resistors.
(a) He wishes to connect some resistors to one of the batteries in such a way to create a circuit that has a current of 0.3333 A leaving the battery. Draw the circuit diagrams for two such circuits below.
(b) He then connect some resistors to his other battery in such a way to create circuit that has a current of 0.6666 A leaving the battery. Draw the circuit diagram for two such circuits below.

2. The current flowing through the 8.45Ω resistor in the circuit below is 1.22 A. What is the voltage of the battery?
Hints: Simplifying your circuit by combining resistors will be advantageous. You should also keep in mind the definitions of what it means for circuit elements to be in series (have the same current passing through them) and parallel (have the same voltage across them).
We should first simplify the circuit by combining all of the circled resistors.
We were told that the 8.45
Ω resistor had a current of 1.22 A, so that is also the current through our 12.6 Ω resistor.
Now the 7.66 Ω resistor and the 12.6 Ω resistor are in parallel, so those two branches of the circuit must have the same voltage drop across them.
V left
=
V right
27.5
Ω
7.66
Ω
12.6
Ω
I R left left
=
I R right right
I left
(
7.66
) (
1.22
A
)(
12.6
Ω )
I left
= 2.01
A
Now we can apply the junction rule to determine the current leaving the battery.
I
Battery
= I
Left
+ I
Right
= 2.01A + 1.22A = 3.23A
So the voltage of the battery must be this current times the effective resistance of the entire circuit.
R eff
= 27.5
V = I R eff
Ω + 4.76 Ω = 32.3 Ω
= (3.23A) (32.3 Ω ) = 104 V