Determine the voltage drop across R2 and R3 in the circuit shown. Take the values to be: E = 24 V, R1 = 500 N, R2 = 200N and R3 = 400 N. (а) 5.1 V R, (b) 12 V E (c) 18.9 V R, R, (d) 24 V
Determine the voltage drop across R2 and R3 in the circuit shown. Take the values to be: E = 24 V, R1 = 500 N, R2 = 200N and R3 = 400 N. (а) 5.1 V R, (b) 12 V E (c) 18.9 V R, R, (d) 24 V
Related questions
Question
100%
![**Problem Statement:**
Determine the voltage drop across \( R_2 \) and \( R_3 \) in the circuit shown. Take the values to be:
\[ E = 24 \, V, \, R_1 = 500 \, \Omega, \, R_2 = 200 \, \Omega \, \text{and} \, R_3 = 400 \, \Omega.\]
**Options:**
(a) 5.1 V
(b) 12 V
(c) 18.9 V
(d) 24 V
**Circuit Diagram Explanation:**
The circuit diagram consists of a series-parallel combination:
- \( E \) represents a 24 V voltage source.
- \( R_1 \), \( R_2 \), and \( R_3 \) are resistors with values 500 Ω, 200 Ω, and 400 Ω respectively.
- \( R_1 \) is in series with the parallel combination of \( R_2 \) and \( R_3 \).
### Steps to Determine the Voltage Drop
1. **Find Equivalent Resistance of the Parallel Combination:**
\[ R_{23} = \left( \frac{1}{R_2} + \frac{1}{R_3} \right)^{-1} = \left( \frac{1}{200} + \frac{1}{400} \right)^{-1} = \left( \frac{1}{200} + \frac{1}{400} \right)^{-1} = \left( \frac{3}{400} \right)^{-1} = \frac{400}{3} \, \Omega \approx 133.33 \, \Omega. \]
2. **Find Total Resistance in the Circuit:**
\[ R_{total} = R_1 + R_{23} = 500 \, \Omega + 133.33 \, \Omega = 633.33 \, \Omega. \]
3. **Find Total Current (I) using Ohm's Law:**
\[ I = \frac{E}{R_{total}} = \frac{24 \, V}{633.33 \, \Omega} \approx 0.0379 \, A. \]
4. **Voltage Drop across the Parallel Combination (V_{23}):**
\[ V_{23} = I \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fabbe649f-9542-4750-a644-5a898c154717%2F8778d7b6-c218-40cf-a125-ea563319c277%2F49ausyb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Determine the voltage drop across \( R_2 \) and \( R_3 \) in the circuit shown. Take the values to be:
\[ E = 24 \, V, \, R_1 = 500 \, \Omega, \, R_2 = 200 \, \Omega \, \text{and} \, R_3 = 400 \, \Omega.\]
**Options:**
(a) 5.1 V
(b) 12 V
(c) 18.9 V
(d) 24 V
**Circuit Diagram Explanation:**
The circuit diagram consists of a series-parallel combination:
- \( E \) represents a 24 V voltage source.
- \( R_1 \), \( R_2 \), and \( R_3 \) are resistors with values 500 Ω, 200 Ω, and 400 Ω respectively.
- \( R_1 \) is in series with the parallel combination of \( R_2 \) and \( R_3 \).
### Steps to Determine the Voltage Drop
1. **Find Equivalent Resistance of the Parallel Combination:**
\[ R_{23} = \left( \frac{1}{R_2} + \frac{1}{R_3} \right)^{-1} = \left( \frac{1}{200} + \frac{1}{400} \right)^{-1} = \left( \frac{1}{200} + \frac{1}{400} \right)^{-1} = \left( \frac{3}{400} \right)^{-1} = \frac{400}{3} \, \Omega \approx 133.33 \, \Omega. \]
2. **Find Total Resistance in the Circuit:**
\[ R_{total} = R_1 + R_{23} = 500 \, \Omega + 133.33 \, \Omega = 633.33 \, \Omega. \]
3. **Find Total Current (I) using Ohm's Law:**
\[ I = \frac{E}{R_{total}} = \frac{24 \, V}{633.33 \, \Omega} \approx 0.0379 \, A. \]
4. **Voltage Drop across the Parallel Combination (V_{23}):**
\[ V_{23} = I \times
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 5 steps
