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Electrical Circuits
Determine the current and voltage for each resistor. Also,
determine the total current coming from the battery.
Similarly, we can consider the loop that starts below the
battery and goes through the 25 Ω resistor to find
the voltage across that resistor.
5.0V  V2  0
5.0V  V2
V=5.0 V
R1 = 35Ω
R2 = 25Ω
V2  5.0V
With the voltage and the resistance, we can calculate the
current.
V  IR
Equation
It is easier for parallel circuits to think of separate loops
for each resistor.
So, first I will look at the loop starting below the battery
then going through the 35 Ω resistor than back to
the bottom of the battery.
5.0V  V1  0
5.0V  V1
V1  5.0V
Using this loop, we can see that the voltage across the
35 Ω resistor is 5 V.
Resistor 1
Solve
Resistor 2
(5V )  I1 (35)
(5V )  I 2 ( 25)
5V
 I1
35
I1  0.143 A
5V
 I2
25
I 2  0 .2 A
The battery must supply both of those currents. So , the
current from the battery is the sum of those two
currents.
I Battery  I1  I 2
Solve
I Battery  (0.143 A)  (0.2 A)
I Battery  0.343 A
The current through resistor 1 is 0.143A and through
resistor 2 is 0.2A, the current through the battery is
0.343A and the voltage across both resistors is 5 V.
Electrical Circuits
Determine the current and the voltage for the following
circuit.
Using the voltage and ohm’s law, we can find the current
through all of the resistors.
V  IR
Equation
V=5.0 V
R1 = 35Ω
R2 = 25Ω
R3 = 15Ω
Looking at the loops. It doesn’t matter which resistor you
consider, you can make a loop that starts at the
bottom of the battery and just goes through that
individual resistor.
Resistor 1
Solve
Resistor 3
(5V )  I1 (35)
(5V )  I 2 ( 25)
(5V )  I 3 (15)
5V
 I1
35
I1  0.143 A
5V
 I2
25
I 2  0 .2 A
5V
 I3
15
I 3  0.33 A
The current from the battery is the total current for all of
the resistors.
5.0V  Vresistor  0
I Battery  I1  I 2  I 3
Vresistor  5.0V
Solve
So the voltage across all of the resistors is 5.0 V.
Resistor 2
I Battery  (0.143 A)  (0.2 A)  (0.33 A)
I Battery  0.673 A
The voltage across the resistors is 5V. The current
through resistor 1 is 0.143A, through resistor 2 is
0.2 A, through resistor 3 is 0.33 A and through the
battery is 0.673 A.
Electrical Circuits
Determine the unknown voltages and currents in the
circuit below.
Now look at the loop that starts below the battery goes
through resistor 1 and then through resistor 3.
9.0V  V1  V3  0
R1 = 10Ω
9.0V  4.5V  V3  0
4.5V  V3  0
R2 = 35Ω
V=9.0 V
V3  4.5V
R3 = 14 Ω
V = 4.5 V
Now using Ohm’s Law we can find the current through
that resistor.
Equation
V  IR
Solve
First, look at the voltage loop that starts below the
battery then goes through resistor 1 and resistor 2.
(4.5V )  I1 (14)
4.5V
 I1
14
I1  0.32 A
9.0V  V1  (4.5V )  0
4.5V  V1  0
V1  4.5V
Using the voltage for resistor 1 and ohm’s law you can
find the current through resistor 1.
Equation
Solve
V  IR
(4.5V )  I1 (10)
4.5V
 I1
10
I1  0.45 A
Finally, looking at resistor 2 using ohm’s law
Equation
V  IR
Solve
(4.5V )  I1 (35)
4.5V
 I1
35
I1  0.13 A
For resistor 1 the voltage drop is 4.5 V and the current is
0.45 A. For resistor 2 the voltage drop is 4.5V and
the current is 0.32 A. For resistor 3 the voltage is
4.5 V and the current is 0.13 A
Electrical Circuits
Determine the unknown voltages and currents for the
following circuit.
Now consider the loop that starts below the battery and
goes through resistor 1 and then resistor 3.
R1 = 20Ω
8.0V  V3  (5V )  0
V=5V
3V  V3  0
V3  3V
R2 = 20Ω
R3 = 30 Ω
Using this voltage and ohm’s law.
Equation
V  IR
V=8.0 V
Solve
(3V )  I 3 (30)
3V
 I3
30
I 3  0.1A
First consider the loop that starts below the battery and
goes the resistor 1 and resistor 2.
8.0V  V2  (5V )  0
Using ohm’s law for resistor 1
Equation
V  IR
3V  V2  0
V2  3V
Using this voltage and ohm’s law.
Equation
V  IR
Solve
(3V )  I 2 ( 20)
3V
 I2
20
I 2  0.15 A
Solve
(5V )  I1 (20)
5V
 I1
20
I1  0.25 A
For resistor 1 the voltage drop is 5 V and the current is
0.25 A. For resistor 2 the voltage drop is 3 V and
the current is 0.15A. For resistor 3 the voltage drop
is 3V and the current is 0.1A.