For the series circuit shown, a) Redraw and label the circuit using standard schematic symbols. b) Draw the direction of the conventional current. flow through the circuit when the switch is closed. c) Calculate the equivalent resistance.

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**Series Circuit Analysis and Calculations**

**For the series circuit shown:**

a) Redraw and label the circuit using standard schematic symbols.

b) Draw the direction of the *conventional current* flow through the circuit when the switch is closed.

c) Calculate the equivalent resistance.

d) Solve for each resistor's I (current), V (voltage), and P (power).

**Circuit Diagram Details:**

The image displays a series circuit consisting of three resistors (1 Ω, 6 Ω, and 2 Ω) connected in series with a 12-Volt battery. There is also a switch present in the circuit. 

**Explanation of the Diagram:**

- The positive terminal of the 12V battery is connected to one terminal of the 1 Ω resistor.
- The other terminal of the 1 Ω resistor is connected to one terminal of the 6 Ω resistor.
- The other terminal of the 6 Ω resistor is connected to one terminal of the 2 Ω resistor.
- The other terminal of the 2 Ω resistor is connected to the switch.
- The switch is then connected back to the negative terminal of the 12V battery.
- This completes the series circuit.

When the switch is closed, conventional current flows from the positive terminal of the battery, through the 1 Ω resistor, then the 6 Ω resistor, followed by the 2 Ω resistor, and finally returns to the battery's negative terminal via the closed switch.

**Instructions for Calculations:**

1. **Equivalent Resistance (R_eq):**

   Since all resistors are in series:
   \[
   R_{eq} = R_1 + R_2 + R_3
   \]
   \[
   R_{eq} = 1 \Omega + 6 \Omega + 2 \Omega = 9 \Omega
   \]

2. **Current (I) in the Circuit:**

   Using Ohm's Law:
   \[
   I = \frac{V}{R_{eq}}
   \]
   \[
   I = \frac{12V}{9 \Omega} = 1.33A
   \]

3. **Voltage (V) across Each Resistor:**

   - For the 1 Ω resistor:
     \[
     V_1 = I \times R_1 = 1.33A \times 1\Omega = 1.33V
     \
Transcribed Image Text:**Series Circuit Analysis and Calculations** **For the series circuit shown:** a) Redraw and label the circuit using standard schematic symbols. b) Draw the direction of the *conventional current* flow through the circuit when the switch is closed. c) Calculate the equivalent resistance. d) Solve for each resistor's I (current), V (voltage), and P (power). **Circuit Diagram Details:** The image displays a series circuit consisting of three resistors (1 Ω, 6 Ω, and 2 Ω) connected in series with a 12-Volt battery. There is also a switch present in the circuit. **Explanation of the Diagram:** - The positive terminal of the 12V battery is connected to one terminal of the 1 Ω resistor. - The other terminal of the 1 Ω resistor is connected to one terminal of the 6 Ω resistor. - The other terminal of the 6 Ω resistor is connected to one terminal of the 2 Ω resistor. - The other terminal of the 2 Ω resistor is connected to the switch. - The switch is then connected back to the negative terminal of the 12V battery. - This completes the series circuit. When the switch is closed, conventional current flows from the positive terminal of the battery, through the 1 Ω resistor, then the 6 Ω resistor, followed by the 2 Ω resistor, and finally returns to the battery's negative terminal via the closed switch. **Instructions for Calculations:** 1. **Equivalent Resistance (R_eq):** Since all resistors are in series: \[ R_{eq} = R_1 + R_2 + R_3 \] \[ R_{eq} = 1 \Omega + 6 \Omega + 2 \Omega = 9 \Omega \] 2. **Current (I) in the Circuit:** Using Ohm's Law: \[ I = \frac{V}{R_{eq}} \] \[ I = \frac{12V}{9 \Omega} = 1.33A \] 3. **Voltage (V) across Each Resistor:** - For the 1 Ω resistor: \[ V_1 = I \times R_1 = 1.33A \times 1\Omega = 1.33V \
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