circuits worksheet.pdf - Free Printable

Problem: Solve the given circuit problems step by step.

#### 1. Calculate the equivalent resistance of the following combination:
The resistors are connected in series:
- Resistances: \( R_1 = 2 \, \Omega \), \( R_2 = 8 \, \Omega \), \( R_3 = 6 \, \Omega \)

For resistors in series, the equivalent resistance \( R_{\text{eq}} \) is given by:
\[
R_{\text{eq}} = R_1 + R_2 + R_3
\]
Substitute the values:
\[
R_{\text{eq}} = 2 + 8 + 6 = 16 \, \Omega
\]

Answer:
\[
\boxed{16 \, \Omega}
\]

---

#### 2. Calculate the equivalent resistance of the following combination:
The resistors are connected in parallel:
- Resistances: \( R_1 = 4 \, \Omega \), \( R_2 = 6 \, \Omega \)

For resistors in parallel, the equivalent resistance \( R_{\text{eq}} \) is given by:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}
\]
Substitute the values:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{6}
\]
Find a common denominator (12):
\[
\frac{1}{R_{\text{eq}}} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}
\]
Thus:
\[
R_{\text{eq}} = \frac{12}{5} = 2.4 \, \Omega
\]

Answer:
\[
\boxed{2.4 \, \Omega}
\]

---

#### 3. Complete the table by calculating the total resistance of the following series circuit. Then calculate the total circuit current and the voltage drops for each of the resistors.

Given:
- Source voltage \( V_{\text{source}} = 12 \, \text{V} \)
- Resistances: \( R_1 = 2.0 \, \Omega \), \( R_2 = 4.0 \, \Omega \), \( R_3 = 6.0 \, \Omega \)

##### Step 1: Calculate the total resistance \( R_{\text{total}} \)
For resistors in series:
\[
R_{\text{total}} = R_1 + R_2 + R_3
\]
Substitute the values:
\[
R_{\text{total}} = 2.0 + 4.0 + 6.0 = 12.0 \, \Omega
\]

##### Step 2: Calculate the total circuit current \( I_{\text{total}} \)
Using Ohm's Law:
\[
I_{\text{total}} = \frac{V_{\text{source}}}{R_{\text{total}}}
\]
Substitute the values:
\[
I_{\text{total}} = \frac{12}{12} = 1.0 \, \text{A}
\]

##### Step 3: Calculate the voltage drop across each resistor
Using Ohm's Law \( V = IR \):

- For \( R_1 \):
\[
V_1 = I_{\text{total}} \cdot R_1 = 1.0 \cdot 2.0 = 2.0 \, \text{V}
\]

- For \( R_2 \):
\[
V_2 = I_{\text{total}} \cdot R_2 = 1.0 \cdot 4.0 = 4.0 \, \text{V}
\]

- For \( R_3 \):
\[
V_3 = I_{\text{total}} \cdot R_3 = 1.0 \cdot 6.0 = 6.0 \, \text{V}
\]

##### Final Table:
\[
\begin{array}{|c|c|c|c|}
\hline
& V & I & R \\
\hline
\text{Source} & 12 \, \text{V} & & \\
\hline
R_1 & 2.0 \, \text{V} & 1.0 \, \text{A} & 2.0 \, \Omega \\
\hline
R_2 & 4.0 \, \text{V} & 1.0 \, \text{A} & 4.0 \, \Omega \\
\hline
R_3 & 6.0 \, \text{V} & 1.0 \, \text{A} & 6.0 \, \Omega \\
\hline
\end{array}
\]

Answer:
\[
\boxed{R_{\text{total}} = 12.0 \, \Omega, \, I_{\text{total}} = 1.0 \, \text{A}, \, V_1 = 2.0 \, \text{V}, \, V_2 = 4.0 \, \text{V}, \, V_3 = 6.0 \, \text{V}}
\]

---

#### 4. Complete the table by calculating the total resistance of the following parallel circuit. Then calculate the total circuit current and the voltage drops and currents for each of the resistors.

Given:
- Source voltage \( V_{\text{source}} = 12 \, \text{V} \)
- Resistances: \( R_1 = 2.0 \, \Omega \), \( R_2 = 3.0 \, \Omega \), \( R_3 = 6.0 \, \Omega \)

##### Step 1: Calculate the total resistance \( R_{\text{total}} \)
For resistors in parallel:
\[
\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]
Substitute the values:
\[
\frac{1}{R_{\text{total}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}
\]
Find a common denominator (6):
\[
\frac{1}{R_{\text{total}}} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1
\]
Thus:
\[
R_{\text{total}} = 1 \, \Omega
\]

##### Step 2: Calculate the total circuit current \( I_{\text{total}} \)
Using Ohm's Law:
\[
I_{\text{total}} = \frac{V_{\text{source}}}{R_{\text{total}}}
\]
Substitute the values:
\[
I_{\text{total}} = \frac{12}{1} = 12 \, \text{A}
\]

##### Step 3: Calculate the current through each resistor
Using Ohm's Law \( I = \frac{V}{R} \):

- For \( R_1 \):
\[
I_1 = \frac{V_{\text{source}}}{R_1} = \frac{12}{2.0} = 6.0 \, \text{A}
\]

- For \( R_2 \):
\[
I_2 = \frac{V_{\text{source}}}{R_2} = \frac{12}{3.0} = 4.0 \, \text{A}
\]

- For \( R_3 \):
\[
I_3 = \frac{V_{\text{source}}}{R_3} = \frac{12}{6.0} = 2.0 \, \text{A}
\]

##### Step 4: Verify the total current
The total current should be the sum of the individual currents:
\[
I_{\text{total}} = I_1 + I_2 + I_3 = 6.0 + 4.0 + 2.0 = 12.0 \, \text{A}
\]
This matches the calculated total current.

##### Final Table:
\[
\begin{array}{|c|c|c|c|}
\hline
& V & I & R \\
\hline
\text{Source} & 12 \, \text{V} & & \\
\hline
R_1 & 12 \, \text{V} & 6.0 \, \text{A} & 2.0 \, \Omega \\
\hline
R_2 & 12 \, \text{V} & 4.0 \, \text{A} & 3.0 \, \Omega \\
\hline
R_3 & 12 \, \text{V} & 2.0 \, \text{A} & 6.0 \, \Omega \\
\hline
\end{array}
\]

Answer:
\[
\boxed{R_{\text{total}} = 1.0 \, \Omega, \, I_{\text{total}} = 12.0 \, \text{A}, \, I_1 = 6.0 \, \text{A}, \, I_2 = 4.0 \, \text{A}, \, I_3 = 2.0 \, \text{A}}
\]

---

#### 5. Calculate the missing information in the table for the following series-parallel network.

Given:
- Source current \( I_{\text{source}} = 2.0 \, \text{A} \)
- Resistances: \( R_1 = 5.0 \, \Omega \), \( R_3 = 4.0 \, \Omega \), \( R_5 = 2.0 \, \Omega \)
- Voltages: \( V_2 = 3.5 \, \text{V} \), \( V_4 = 1.5 \, \text{V} \)

##### Step 1: Calculate the voltage across \( R_1 \)
Using Ohm's Law:
\[
V_1 = I_{\text{source}} \cdot R_1
\]
Substitute the values:
\[
V_1 = 2.0 \cdot 5.0 = 10.0 \, \text{V}
\]

##### Step 2: Calculate the voltage across \( R_3 \)
Using Ohm's Law:
\[
V_3 = I_{\text{source}} \cdot R_3
\]
Substitute the values:
\[
V_3 = 2.0 \cdot 4.0 = 8.0 \, \text{V}
\]

##### Step 3: Calculate the voltage across \( R_5 \)
Using Ohm's Law:
\[
V_5 = I_{\text{source}} \cdot R_5
\]
Substitute the values:
\[
V_5 = 2.0 \cdot 2.0 = 4.0 \, \text{V}
\]

##### Step 4: Calculate the total source voltage
The total source voltage is the sum of the voltages across all resistors:
\[
V_{\text{source}} = V_1 + V_2 + V_3 + V_4 + V_5
\]
Substitute the known values:
\[
V_{\text{source}} = 10.0 + 3.5 + 8.0 + 1.5 + 4.0 = 27.0 \, \text{V}
\]

##### Step 5: Calculate the resistance \( R_2 \)
Using Ohm's Law:
\[
R_2 = \frac{V_2}{I_{\text{source}}}
\]
Substitute the values:
\[
R_2 = \frac{3.5}{2.0} = 1.75 \, \Omega
\]

##### Step 6: Calculate the resistance \( R_4 \)
Using Ohm's Law:
\[
R_4 = \frac{V_4}{I_{\text{source}}}
\]
Substitute the values:
\[
R_4 = \frac{1.5}{2.0} = 0.75 \, \Omega
\]

##### Final Table:
\[
\begin{array}{|c|c|c|c|}
\hline
& V & I & R \\
\hline
\text{Source} & 27.0 \, \text{V} & 2.0 \, \text{A} & \\
\hline
R_1 & 10.0 \, \text{V} & 2.0 \, \text{A} & 5.0 \, \Omega \\
\hline
R_2 & 3.5 \, \text{V} & 2.0 \, \text{A} & 1.75 \, \Omega \\
\hline
R_3 & 8.0 \, \text{V} & 2.0 \, \text{A} & 4.0 \, \Omega \\
\hline
R_4 & 1.5 \, \text{V} & 2.0 \, \text{A} & 0.75 \, \Omega \\
\hline
R_5 & 4.0 \, \text{V} & 2.0 \, \text{A} & 2.0 \, \Omega \\
\hline
\end{array}
\]

Answer:
\[
\boxed{V_{\text{source}} = 27.0 \, \text{V}, \, R_2 = 1.75 \, \Omega, \, R_4 = 0.75 \, \Omega, \, V_1 = 10.0 \, \text{V}, \, V_3 = 8.0 \, \text{V}, \, V_5 = 4.0 \, \text{V}}
\]