21. What is the voltage drop across the 20-ohm resistor in the circuit? A. 6 V B. 4 V C. 2 V D. OV

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Chapter10: Direct-current Circuits
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Problem 72AP: The rather simple circuit shown below is known as a voltage divider. The symbol consisting of three...
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**Question 21:** What is the voltage drop across the 20-ohm resistor in the circuit?

**Options:**
- A. 6 V
- B. 4 V
- C. 2 V
- D. 0 V

**Circuit Diagram Explanation:**

The diagram shows a circuit with a 6V battery connected across a parallel combination of two resistors, each with a resistance of 10 ohms. This parallel section is then connected in series with a 20-ohm resistor.

To solve for the voltage drop across the 20-ohm resistor, we need to consider the total equivalent resistance of the circuit and the voltage distribution according to Ohm's Law.

1. **Find equivalent resistance of the parallel section**:
   \[
   \frac{1}{R_{\text{parallel}}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} \quad \Rightarrow \quad R_{\text{parallel}} = 5 \, \Omega
   \]

2. **Find total resistance in the circuit**:
   \[
   R_{\text{total}} = R_{\text{parallel}} + R_{20 \, \Omega} = 5 \, \Omega + 20 \, \Omega = 25 \, \Omega
   \]

3. **Calculate current using Ohm's Law**:
   \[
   I = \frac{V}{R_{\text{total}}} = \frac{6 \, \text{V}}{25 \, \Omega} = 0.24 \, \text{A}
   \]

4. **Calculate voltage drop across the 20-ohm resistor**:
   \[
   V_{\text{20} \, \Omega} = I \times R_{20 \, \Omega} = 0.24 \, \text{A} \times 20 \, \Omega = 4.8 \, \text{V}
   \]

Therefore, the closest correct answer is:
- **B. 4 V**
Transcribed Image Text:**Question 21:** What is the voltage drop across the 20-ohm resistor in the circuit? **Options:** - A. 6 V - B. 4 V - C. 2 V - D. 0 V **Circuit Diagram Explanation:** The diagram shows a circuit with a 6V battery connected across a parallel combination of two resistors, each with a resistance of 10 ohms. This parallel section is then connected in series with a 20-ohm resistor. To solve for the voltage drop across the 20-ohm resistor, we need to consider the total equivalent resistance of the circuit and the voltage distribution according to Ohm's Law. 1. **Find equivalent resistance of the parallel section**: \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} \quad \Rightarrow \quad R_{\text{parallel}} = 5 \, \Omega \] 2. **Find total resistance in the circuit**: \[ R_{\text{total}} = R_{\text{parallel}} + R_{20 \, \Omega} = 5 \, \Omega + 20 \, \Omega = 25 \, \Omega \] 3. **Calculate current using Ohm's Law**: \[ I = \frac{V}{R_{\text{total}}} = \frac{6 \, \text{V}}{25 \, \Omega} = 0.24 \, \text{A} \] 4. **Calculate voltage drop across the 20-ohm resistor**: \[ V_{\text{20} \, \Omega} = I \times R_{20 \, \Omega} = 0.24 \, \text{A} \times 20 \, \Omega = 4.8 \, \text{V} \] Therefore, the closest correct answer is: - **B. 4 V**
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