Question
Look at the following series circuit
12v on the left side 1 inch opening
3 n in the center 3 Triangles
5n on the right side 3 Triangles
7n on the right conor 3 Triangles
Calculate the voltage drop across each resistor, R1, R2 and R3. Show all work and calculations
*Hint - Use equivalent resistance to find the current across each resistor first
12v on the left side 1 inch opening
3 n in the center 3 Triangles
5n on the right side 3 Triangles
7n on the right conor 3 Triangles
Calculate the voltage drop across each resistor, R1, R2 and R3. Show all work and calculations
*Hint - Use equivalent resistance to find the current across each resistor first
Answers
Answered by
GPT-4o mini
To calculate the voltage drop across each resistor in the series circuit you've described, we first need to clarify the values of the resistors (R1, R2, and R3) and their configuration in the circuit.
Based on your description:
- The "12V" indicates the total voltage across the circuit.
- The notations "3 n" and "5 n" seem to indicate resistance values (in ohms).
- "7 n on the right corner" might suggest an additional resistor (R3).
Let's assume your resistors are as follows:
- R1 = 3 ohms (from "3 n in the center")
- R2 = 5 ohms (from "5 n on the right side")
- R3 = 7 ohms (from "7 n on the right corner")
**Step 1: Calculate the Equivalent Resistance (R_eq)**
In a series circuit, the total or equivalent resistance (R_eq) is simply the sum of all resistances:
\[
R_{eq} = R_1 + R_2 + R_3 = 3 \, \Omega + 5 \, \Omega + 7 \, \Omega = 15 \, \Omega
\]
**Step 2: Find the Total Current (I)**
Using Ohm’s law (\(V = IR\)), we can find the current flowing through the circuit:
\[
I = \frac{V}{R_{eq}} = \frac{12 \, V}{15 \, \Omega} = 0.8 \, A
\]
**Step 3: Calculate the Voltage Drop Across Each Resistor**
The voltage drop across a resistor in a series circuit is given by \(V = IR\). We can calculate the voltage drop for each resistor:
- **For R1 (3 ohms)**:
\[
V_{R1} = I \times R_1 = 0.8 \, A \times 3 \, \Omega = 2.4 \, V
\]
- **For R2 (5 ohms)**:
\[
V_{R2} = I \times R_2 = 0.8 \, A \times 5 \, \Omega = 4 \, V
\]
- **For R3 (7 ohms)**:
\[
V_{R3} = I \times R_3 = 0.8 \, A \times 7 \, \Omega = 5.6 \, V
\]
**Step 4: Verify the Total Voltage**
Now, let’s add all the voltage drops together to ensure they equal the total supplied voltage (12V):
\[
V_{R1} + V_{R2} + V_{R3} = 2.4 \, V + 4 \, V + 5.6 \, V = 12 \, V
\]
**Final Results**:
- Voltage drop across R1: \(2.4 \, V\)
- Voltage drop across R2: \(4 \, V\)
- Voltage drop across R3: \(5.6 \, V\)
This confirms that our calculations are correct, and the voltage drops across the resistors in the series circuit are as calculated.
Based on your description:
- The "12V" indicates the total voltage across the circuit.
- The notations "3 n" and "5 n" seem to indicate resistance values (in ohms).
- "7 n on the right corner" might suggest an additional resistor (R3).
Let's assume your resistors are as follows:
- R1 = 3 ohms (from "3 n in the center")
- R2 = 5 ohms (from "5 n on the right side")
- R3 = 7 ohms (from "7 n on the right corner")
**Step 1: Calculate the Equivalent Resistance (R_eq)**
In a series circuit, the total or equivalent resistance (R_eq) is simply the sum of all resistances:
\[
R_{eq} = R_1 + R_2 + R_3 = 3 \, \Omega + 5 \, \Omega + 7 \, \Omega = 15 \, \Omega
\]
**Step 2: Find the Total Current (I)**
Using Ohm’s law (\(V = IR\)), we can find the current flowing through the circuit:
\[
I = \frac{V}{R_{eq}} = \frac{12 \, V}{15 \, \Omega} = 0.8 \, A
\]
**Step 3: Calculate the Voltage Drop Across Each Resistor**
The voltage drop across a resistor in a series circuit is given by \(V = IR\). We can calculate the voltage drop for each resistor:
- **For R1 (3 ohms)**:
\[
V_{R1} = I \times R_1 = 0.8 \, A \times 3 \, \Omega = 2.4 \, V
\]
- **For R2 (5 ohms)**:
\[
V_{R2} = I \times R_2 = 0.8 \, A \times 5 \, \Omega = 4 \, V
\]
- **For R3 (7 ohms)**:
\[
V_{R3} = I \times R_3 = 0.8 \, A \times 7 \, \Omega = 5.6 \, V
\]
**Step 4: Verify the Total Voltage**
Now, let’s add all the voltage drops together to ensure they equal the total supplied voltage (12V):
\[
V_{R1} + V_{R2} + V_{R3} = 2.4 \, V + 4 \, V + 5.6 \, V = 12 \, V
\]
**Final Results**:
- Voltage drop across R1: \(2.4 \, V\)
- Voltage drop across R2: \(4 \, V\)
- Voltage drop across R3: \(5.6 \, V\)
This confirms that our calculations are correct, and the voltage drops across the resistors in the series circuit are as calculated.