Resistors Circuit

## **Problem**

Determine the voltage, current, resistance, and power for each resister in the circuit below.

## **Solution**

Construct a **VIRP table** to keep track of the voltage, current, resistance, and power of all the parts of the circuit. The units for voltage, current, resistance, and power are volts, amps, ohms, and joules respectively. Fill in what is already known.

The next step is to solve for the total resistance of the entire circuit. Often times this task is easier said than done (especially when there are more than four resistors). The end goal is to combine all the resistors into one equivalent resistor with the total resistance

R
T

{\displaystyle R\_T}

. To achieve this goal we should work towards making the resistors add in series. For the above circuit, we need to find the equivalent resistor

R

1234

{\displaystyle R\_{1234}}

.

Begin by combining the parallel resistors

R
3

{\displaystyle R\_3 }

and

R
4

{\displaystyle R\_4}

.

1

R

34

=

1

R
3

1

R
4

1

R

34

=

1
8

1
4

R

34

=

8
3

Ω

{\displaystyle \begin{align}
\frac{1}{R\_{34}} &= \frac{1}{R\_3} + \frac{1}{R\_4} \\[5 pt]
\frac{1}{R\_{34}} &= \frac{1}{8} + \frac{1}{4}\\[5 pt]
R\_{34} &= \frac{8}{3} \, \Omega
\end{align}}

Next combine the resistors

R
2

{\displaystyle R\_2}

 and

R

34

{\displaystyle R\_{34}}

 in series.

R

234

=

R
2

R

34

R

234

=
6

8
3

R

234

=

26
3

Ω

{\displaystyle \begin{align}
R\_{234} &= R\_2 + R\_{34} \\[5 pt]
R\_{234} &= 6 + \frac{8}{3}\\[5 pt]
R\_{234} &= \frac{26}{3} \, \Omega
\end{align}}

Next combine the parallel resistors

R
1

{\displaystyle R\_1}

 and

R

234

{\displaystyle R\_{234}}

.

1

R

1234

=

1

R
1

1

R

234

1

R

1234

=

1
10

3
26

R

1234

=

65
14

Ω

{\displaystyle \begin{align}
\frac{1}{R\_{1234}} &= \frac{1}{R\_1} + \frac{1}{R\_{234}} \\[5 pt]
\frac{1}{R\_{1234}} &= \frac{1}{10} + \frac{3}{26}\\[5 pt]
R\_{1234} &= \frac{65}{14} \, \Omega
\end{align}}

Now that the equivalent resistance

R

1234

{\displaystyle R\_{1234}}

 is found, we can add this result in series with

# r
0.5

Ω

{\displaystyle r = 0.5 \; \Omega}

 and

R
5

=
5

Ω

{\displaystyle R\_5 = 5 \; \Omega}

 to obtain

R
T

{\displaystyle R\_T}

.

R
T

=
r

R

1234

R
5

R
T

=
0.5

65
14

5

R
T

=

71
7

Ω

{\displaystyle \begin{align}
R\_T &= r + R\_{1234} + R\_5 \\[5 pt]
R\_T &= 0.5 + \frac{65}{14} + 5\\[5 pt]
R\_T &= \frac{71}{7} \, \Omega
\end{align}}

With the total resistance, we can determine the current of the total circuit.

9

=

I
T

(

71
7

)

I
T

=

63
71

A

{\displaystyle \begin{align}
9 &= I\_T \left(\frac{71}{7} \right) \\[5 pt]
I\_T &= \frac{63}{71} \; A
\end{align}}

This current is also the current of the resistor

r

{\displaystyle r}

 and

R
5

{\displaystyle R\_5}

; thus, their respective voltages can be found.

V
r

=

63
71

⋅
# 0.5

63
142

V

V

R
5

=

63
71

⋅
# 5

315
71

V

{\displaystyle \begin{align}
V\_r &= \frac{63}{71} \cdot 0.5 = \frac{63}{142}\; V \\[5 pt]
V\_{R\_5} &= \frac{63}{71} \cdot 5 = \frac{315}{71} \; V
\end{align}}

Update the VIRP table.

Now it is time to analyze the resistors that make up

R

1234

{\displaystyle R\_{1234}}

. The voltage across

R
1

{\displaystyle R\_1}

 and

R

234

{\displaystyle R\_{234}}

 are equal to

V

1234

{\displaystyle V\_{1234}}

 because they are in parallel. By **Kirchoff's loop rule**, we get

E

−
r
I
−

V

1234

−

R
5

# I
0

V

1234

=

E

−
(
r

R
5

)
I

V

1234

=
9
−
(
0.5
5
)

63
71

=

585
142

V

{\displaystyle \begin{align}
\mathcal{E} - rI - V\_{1234} - R\_5I = 0 \\[5 pt]
V\_{1234} = \mathcal{E} - (r + R\_5) I \\[5 pt]
V\_{1234} = 9 - (0.5 + 5) \frac{63}{71} = \frac{585}{142} \; V
\end{align}}

Therefore,

V

R
1

=

585
142

V

{\displaystyle V\_{R\_1} = \frac{585}{142} \; V}

 and the current through resistor

R
1

{\displaystyle R\_1}

 is

V

R
1

=

I

R
1

R
1

585
142

=

I

R
1

⋅
10

R
1

=

117
284

A

{\displaystyle \begin{align}
V\_{R\_1} &= I\_{R\_1} R\_1 \\[5 pt]
\frac{585}{142} &= I\_{R\_1} \cdot 10 \\[5 pt]
R\_1 &= \frac{117}{284} \; A
\end{align}}

By **Kirchoff's junction rule**, the current through

R
2

{\displaystyle R\_2}

 must be

I

R
2

=

I
T

−

I

R
1

I

R
2

=

63
71

−

117
284

I

R
2

=

135
284

A

{\displaystyle \begin{align}
I\_{R\_2} &= I\_T - I\_{R\_1} \\[5 pt]
I\_{R\_2} &= \frac{63}{71} - \frac{117}{284} \\[5 pt]
I\_{R\_2} &= \frac{135}{284} \; A
\end{align}}

Using this current we can determine the voltage across resistor

R
2

{\displaystyle R\_2}

.

V

R
2

=

I

R
2

R
2

V

R
2

=

135
284

⋅
6

V

R
2

=

405
142

V

{\displaystyle \begin{align}
V\_{R\_2} &= I\_{R\_2} R\_2 \\[5 pt]
V\_{R\_2} &= \frac{135}{284}\cdot 6 \\[5 pt]
V\_{R\_2} &= \frac{405}{142} \; V
\end{align}}

Update the VIRP table.

The voltages for resistors

R
3

{\displaystyle R\_3 }

 and

R
4

{\displaystyle R\_4}

 are equal because they are in parallel, and can be determined by subtracting the voltages of resistors

R
1

{\displaystyle R\_1}

 and

R
2

{\displaystyle R\_2}

.

V

R
3

=

V

R
1

−

V

R
2

V

R
3

=

585
142

−

405
142

V

R
3

=

90
71

V

{\displaystyle \begin{align}
V\_{R\_3} &= V\_{R\_1} - V\_{R\_2} \\[5 pt]
V\_{R\_3} &= \frac{585}{142} - \frac{405}{142} \\[5 pt]
V\_{R\_3} &= \frac{90}{71} \; V
\end{align}}

Dividing this voltage by the resistance of

R
3

{\displaystyle R\_3 }

 and

R
4

{\displaystyle R\_4}

 to get their respective currents.

V

R
3

=

I

R
3

R
3

90
71

=

I

R
3

⋅
8

R
3

=

45
284

A

{\displaystyle \begin{align}
V\_{R\_3} &= I\_{R\_3} R\_3 \\[5 pt]
\frac{90}{71} &= I\_{R\_3} \cdot 8 \\[5 pt]
R\_3 &= \frac{45}{284} \; A
\end{align}}

V

R
4

=

I

R
4

R
4

90
71

=

I

R
4

⋅
4

R
4

=

45
142

A

{\displaystyle \begin{align}
V\_{R\_4} &= I\_{R\_4} R\_4 \\[5 pt]
\frac{90}{71} &= I\_{R\_4} \cdot 4 \\[5 pt]
R\_4 &= \frac{45}{142} \; A
\end{align}}

Update the VIRP table.

Convert all the values to decimals (to four significant figures here). Use the formula

# P
I
V

{\displaystyle P=IV}

 to calculate the power dissipated by each resistor and the entire circuit.