A 12.0V battery is wired in series with 3 resistors, 8 Ohm, 15 Ohm and 75 Ohm. a.) Using either the computer drawing tools or a scanned hand written diagram, draw this circuit including proper symbols and labels. (Do not copy and paste an image from any other resource) b.) Find the equivalent resistance and current flowing through each resistor. Show your work and label your diagram with the current amount and direction of flow. GEind the voltage drop, across each resistor in the circuit. Show your work.

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### Example Problem on Series Circuit

**Problem:**

A 12.0V battery is wired in series with 3 resistors, 8 Ohm, 15 Ohm, and 75 Ohm.

**Questions:**

a.) Using either computer drawing tools or a scanned handwritten diagram, draw this circuit including proper symbols and labels. (Do not copy and paste an image from any other resource)

b.) Find the equivalent resistance and current flowing through each resistor. Show your work and label your diagram with the current amount and direction of flow.

c.) Find the voltage drop across each resistor in the circuit. Show your work.

### Solution:

#### a.) Circuit Diagram
(Students are required to draw a simple series circuit with a 12V battery and three resistors labeled as 8Ω, 15Ω, and 75Ω. Ensure you include the proper connection points and direction of current flow.)

#### b.) Finding Equivalent Resistance and Current:

Step 1: Calculate the equivalent resistance (R_eq) in a series circuit.
\[ R_{eq} = R_1 + R_2 + R_3 \]
\[ R_{eq} = 8 \Omega + 15 \Omega + 75 \Omega \]
\[ R_{eq} = 98 \Omega \]

Step 2: Calculate the current (I) using Ohm's Law.
\[ V = IR \]
\[ I = \frac{V}{R} = \frac{12V}{98 \Omega} \approx 0.1224 \, A \]

So, the current flowing through each resistor is approximately 0.1224 A (since the resistors are in series, the current is the same through each resistor).

#### c.) Finding Voltage Drop Across Each Resistor:

Using Ohm's Law, \( V = IR \),

- **Voltage drop across 8Ω resistor (V1):**
  \[ V_1 = I \times R_1 \]
  \[ V_1 = 0.1224A \times 8 \Omega \]
  \[ V_1 \approx 0.9792V \]

- **Voltage drop across 15Ω resistor (V2):**
  \[ V_2 = I \times R_2 \]
  \[ V_2 = 0.1224A \times 15 \Omega \]
  \[ V_2 \
Transcribed Image Text:### Example Problem on Series Circuit **Problem:** A 12.0V battery is wired in series with 3 resistors, 8 Ohm, 15 Ohm, and 75 Ohm. **Questions:** a.) Using either computer drawing tools or a scanned handwritten diagram, draw this circuit including proper symbols and labels. (Do not copy and paste an image from any other resource) b.) Find the equivalent resistance and current flowing through each resistor. Show your work and label your diagram with the current amount and direction of flow. c.) Find the voltage drop across each resistor in the circuit. Show your work. ### Solution: #### a.) Circuit Diagram (Students are required to draw a simple series circuit with a 12V battery and three resistors labeled as 8Ω, 15Ω, and 75Ω. Ensure you include the proper connection points and direction of current flow.) #### b.) Finding Equivalent Resistance and Current: Step 1: Calculate the equivalent resistance (R_eq) in a series circuit. \[ R_{eq} = R_1 + R_2 + R_3 \] \[ R_{eq} = 8 \Omega + 15 \Omega + 75 \Omega \] \[ R_{eq} = 98 \Omega \] Step 2: Calculate the current (I) using Ohm's Law. \[ V = IR \] \[ I = \frac{V}{R} = \frac{12V}{98 \Omega} \approx 0.1224 \, A \] So, the current flowing through each resistor is approximately 0.1224 A (since the resistors are in series, the current is the same through each resistor). #### c.) Finding Voltage Drop Across Each Resistor: Using Ohm's Law, \( V = IR \), - **Voltage drop across 8Ω resistor (V1):** \[ V_1 = I \times R_1 \] \[ V_1 = 0.1224A \times 8 \Omega \] \[ V_1 \approx 0.9792V \] - **Voltage drop across 15Ω resistor (V2):** \[ V_2 = I \times R_2 \] \[ V_2 = 0.1224A \times 15 \Omega \] \[ V_2 \
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