Series And Parallel Circuits Worksheet Answer Key - Free Printable

Problem Description:

The task involves analyzing two series-parallel circuits to calculate the total resistance and then use Ohm's Law to determine the total current supplied by the power supply. The circuits are provided in the image, and hints are given for simplifying the calculations.

#### Circuit 1:
- Resistors: \( R_1 = 390 \, \Omega \), \( R_2 = 820 \, \Omega \), \( R_3 = 330 \, \Omega \), \( R_4 = 220 \, \Omega \), \( R_5 = 270 \, \Omega \), \( R_6 = 1.2 \, \text{k}\Omega \)
- Voltage source: \( V_T = 50 \, \text{V} \)

#### Circuit 2:
- Resistors: \( R_1 = 1 \, \text{k}\Omega \), \( R_2 = 1.2 \, \text{k}\Omega \), \( R_3 = 1.2 \, \text{k}\Omega \), \( R_4 = 100 \, \Omega \), \( R_5 = 120 \, \Omega \), \( R_6 = 18 \, \Omega \), \( R_7 = 12 \, \Omega \), \( R_8 = 270 \, \Omega \), \( R_9 = 10 \, \Omega \)
- Voltage source: \( V_T = 50 \, \text{V} \)

Solution:

#### Step 1: Simplify Circuit 1

##### Circuit Diagram:
 (Insert Circuit 1 diagram here)

##### Step 1.1: Combine Parallel Resistors \( R_3 \) and \( R_4 \):
Resistors \( R_3 \) and \( R_4 \) are in parallel. The equivalent resistance \( R_{34} \) is calculated as:
\[
\frac{1}{R_{34}} = \frac{1}{R_3} + \frac{1}{R_4}
\]
\[
\frac{1}{R_{34}} = \frac{1}{330} + \frac{1}{220}
\]
\[
\frac{1}{R_{34}} = \frac{220 + 330}{330 \times 220} = \frac{550}{72600}
\]
\[
R_{34} = \frac{72600}{550} = 132 \, \Omega
\]

##### Step 1.2: Combine Series Resistors \( R_2 \) and \( R_{34} \):
Resistors \( R_2 \) and \( R_{34} \) are in series. The equivalent resistance \( R_{234} \) is:
\[
R_{234} = R_2 + R_{34}
\]
\[
R_{234} = 820 + 132 = 952 \, \Omega
\]

##### Step 1.3: Combine Parallel Resistors \( R_5 \) and \( R_{234} \):
Resistors \( R_5 \) and \( R_{234} \) are in parallel. The equivalent resistance \( R_{5234} \) is:
\[
\frac{1}{R_{5234}} = \frac{1}{R_5} + \frac{1}{R_{234}}
\]
\[
\frac{1}{R_{5234}} = \frac{1}{270} + \frac{1}{952}
\]
\[
\frac{1}{R_{5234}} = \frac{952 + 270}{270 \times 952} = \frac{1222}{257040}
\]
\[
R_{5234} = \frac{257040}{1222} \approx 210.34 \, \Omega
\]

##### Step 1.4: Combine Series Resistors \( R_1 \), \( R_6 \), and \( R_{5234} \):
Resistors \( R_1 \), \( R_6 \), and \( R_{5234} \) are in series. The total resistance \( R_{\text{total}} \) is:
\[
R_{\text{total}} = R_1 + R_6 + R_{5234}
\]
\[
R_{\text{total}} = 390 + 1200 + 210.34 = 1800.34 \, \Omega
\]

##### Step 1.5: Calculate Total Current Using Ohm's Law:
Using Ohm's Law \( I = \frac{V}{R} \):
\[
I_{\text{total}} = \frac{V_T}{R_{\text{total}}}
\]
\[
I_{\text{total}} = \frac{50}{1800.34} \approx 0.0278 \, \text{A} = 27.8 \, \text{mA}
\]

#### Final Answer for Circuit 1:
\[
\boxed{1800.34 \, \Omega, 27.8 \, \text{mA}}
\]

#### Step 2: Simplify Circuit 2

##### Circuit Diagram:
 (Insert Circuit 2 diagram here)

##### Step 2.1: Combine Parallel Resistors \( R_7 \), \( R_8 \), and \( R_9 \):
Resistors \( R_7 \), \( R_8 \), and \( R_9 \) are in parallel. The equivalent resistance \( R_{789} \) is:
\[
\frac{1}{R_{789}} = \frac{1}{R_7} + \frac{1}{R_8} + \frac{1}{R_9}
\]
\[
\frac{1}{R_{789}} = \frac{1}{12} + \frac{1}{270} + \frac{1}{10}
\]
\[
\frac{1}{R_{789}} = \frac{270 + 12 + 324}{3240} = \frac{606}{3240}
\]
\[
R_{789} = \frac{3240}{606} \approx 5.35 \, \Omega
\]

##### Step 2.2: Combine Series Resistors \( R_5 \) and \( R_{789} \):
Resistors \( R_5 \) and \( R_{789} \) are in series. The equivalent resistance \( R_{5789} \) is:
\[
R_{5789} = R_5 + R_{789}
\]
\[
R_{5789} = 120 + 5.35 = 125.35 \, \Omega
\]

##### Step 2.3: Combine Parallel Resistors \( R_4 \) and \( R_{5789} \):
Resistors \( R_4 \) and \( R_{5789} \) are in parallel. The equivalent resistance \( R_{45789} \) is:
\[
\frac{1}{R_{45789}} = \frac{1}{R_4} + \frac{1}{R_{5789}}
\]
\[
\frac{1}{R_{45789}} = \frac{1}{100} + \frac{1}{125.35}
\]
\[
\frac{1}{R_{45789}} = \frac{125.35 + 100}{12535} = \frac{225.35}{12535}
\]
\[
R_{45789} = \frac{12535}{225.35} \approx 55.6 \, \Omega
\]

##### Step 2.4: Combine Series Resistors \( R_1 \), \( R_2 \), \( R_3 \), and \( R_{45789} \):
Resistors \( R_1 \), \( R_2 \), \( R_3 \), and \( R_{45789} \) are in series. The total resistance \( R_{\text{total}} \) is:
\[
R_{\text{total}} = R_1 + R_2 + R_3 + R_{45789}
\]
\[
R_{\text{total}} = 1000 + 1200 + 1200 + 55.6 = 3455.6 \, \Omega
\]

##### Step 2.5: Calculate Total Current Using Ohm's Law:
Using Ohm's Law \( I = \frac{V}{R} \):
\[
I_{\text{total}} = \frac{V_T}{R_{\text{total}}}
\]
\[
I_{\text{total}} = \frac{50}{3455.6} \approx 0.0145 \, \text{A} = 14.5 \, \text{mA}
\]

#### Final Answer for Circuit 2:
\[
\boxed{3455.6 \, \Omega, 14.5 \, \text{mA}}
\]

Summary of Results:

1. Circuit 1:
- Total Resistance: \( 1800.34 \, \Omega \)
- Total Current: \( 27.8 \, \text{mA} \)

2. Circuit 2:
- Total Resistance: \( 3455.6 \, \Omega \)
- Total Current: \( 14.5 \, \text{mA} \)