or the below, the wn voltage drops across resistors R1 and R4: Provide your calculations and reasoning for your answer.) R1 R3 + VR3 = 2.5 V V1 + + 25Vdc R2 R4 VR2 = 9.75 V
or the below, the wn voltage drops across resistors R1 and R4: Provide your calculations and reasoning for your answer.) R1 R3 + VR3 = 2.5 V V1 + + 25Vdc R2 R4 VR2 = 9.75 V
Introductory Circuit Analysis (13th Edition)
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Publisher:Robert L. Boylestad
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Question
![**Circuit Analysis Problem**
For the circuit below, determine the unknown voltage drops across resistors R1 and R4. Provide your calculations and reasoning for your answer.
**Circuit Diagram Explanation**
The circuit is a simple series-parallel configuration with a 25V DC voltage source (V1) on the left. R1 is directly in series with the voltage source, while R2 is in parallel with R3 and R4 is positioned further along the series path after R3. The following details are provided:
- Voltage across R2 (\(V_{R2}\)) is 9.75 V.
- Voltage across R3 (\(V_{R3}\)) is 2.5 V.
**Steps to Solve**
1. **Total Voltage**: The total voltage of the circuit provided by V1 is 25V.
2. **Parallel Voltage**: Since R2 and R3 are in parallel, both resistors share the same voltage across them which is 9.75 V. Verify this from the problem as \(V_{R2}\) has been given.
3. **Voltage Division**: The voltage drop across resistors in series can be found using the rule that the sum of voltage drops in a closed loop equals the total voltage supplied.
4. **Voltage Across R4**:
- By Kirchhoff's Voltage Law:
\[
V_{R4} = V1 - (V_{R3} + V_{R2})
\]
Calculate \(V_{R4}\):
\[
V_{R4} = 25 - (2.5 + 9.75) = 25 - 12.25 = 12.75 \text{ V}
\]
5. **Voltage Across R1**:
- Since \(V_{R1} = V1 - V_{R2}\)
Calculate \(V_{R1}\):
\[
V_{R1} = 25 - 9.75 = 15.25 \text{ V}
\]
These calculations give you the voltage drops across the unknown resistors R1 and R4 in the circuit.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07546766-63d0-4d74-b8c6-8a4c09c9f342%2F7c26b989-d227-4d00-9453-fea8aa9d3a00%2F70uypv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Circuit Analysis Problem**
For the circuit below, determine the unknown voltage drops across resistors R1 and R4. Provide your calculations and reasoning for your answer.
**Circuit Diagram Explanation**
The circuit is a simple series-parallel configuration with a 25V DC voltage source (V1) on the left. R1 is directly in series with the voltage source, while R2 is in parallel with R3 and R4 is positioned further along the series path after R3. The following details are provided:
- Voltage across R2 (\(V_{R2}\)) is 9.75 V.
- Voltage across R3 (\(V_{R3}\)) is 2.5 V.
**Steps to Solve**
1. **Total Voltage**: The total voltage of the circuit provided by V1 is 25V.
2. **Parallel Voltage**: Since R2 and R3 are in parallel, both resistors share the same voltage across them which is 9.75 V. Verify this from the problem as \(V_{R2}\) has been given.
3. **Voltage Division**: The voltage drop across resistors in series can be found using the rule that the sum of voltage drops in a closed loop equals the total voltage supplied.
4. **Voltage Across R4**:
- By Kirchhoff's Voltage Law:
\[
V_{R4} = V1 - (V_{R3} + V_{R2})
\]
Calculate \(V_{R4}\):
\[
V_{R4} = 25 - (2.5 + 9.75) = 25 - 12.25 = 12.75 \text{ V}
\]
5. **Voltage Across R1**:
- Since \(V_{R1} = V1 - V_{R2}\)
Calculate \(V_{R1}\):
\[
V_{R1} = 25 - 9.75 = 15.25 \text{ V}
\]
These calculations give you the voltage drops across the unknown resistors R1 and R4 in the circuit.
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