For circuit series and parallel, the supply voltage of, Vsupply=240 Volts, R1= 202, R2=302, R3=82, R4=302 and R5= 602. What will be the total resistor value? Vsupply 꿈 {R₂
For circuit series and parallel, the supply voltage of, Vsupply=240 Volts, R1= 202, R2=302, R3=82, R4=302 and R5= 602. What will be the total resistor value? Vsupply 꿈 {R₂
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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KVL and KCL
KVL stands for Kirchhoff voltage law. KVL states that the total voltage drops around the loop in any closed electric circuit is equal to the sum of total voltage drop in the same closed loop.
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Question
![**Problem Statement:**
For circuits in series and parallel, the supply voltage \(V_{\text{Supply}} = 240\) Volts. The resistor values are as follows:
- \(R_1 = 20 \, \Omega\)
- \(R_2 = 30 \, \Omega\)
- \(R_3 = 8 \, \Omega\)
- \(R_4 = 30 \, \Omega\)
- \(R_5 = 60 \, \Omega\)
What will be the total resistance value?
**Diagram Explanation:**
The diagram illustrates a circuit with a combination of series and parallel resistors:
- \(R_1\) and \(R_2\) are in parallel at the top part of the circuit.
- \(R_3\) is in series with the parallel combination of \(R_4\) and \(R_5\).
- \(R_4\) and \(R_5\) are in parallel at the bottom of the circuit.
**Answer Choices:**
- \(40 \, \Omega\)
- \(20 \, \Omega\)
- \(100 \, \Omega\)
- \(60 \, \Omega\)
**Solution Explanation:**
To find the total resistance:
1. Calculate the equivalent resistance of \(R_1\) and \(R_2\) in parallel:
\[
\frac{1}{R_{\text{parallel1}}} = \frac{1}{R_1} + \frac{1}{R_2}
= \frac{1}{20} + \frac{1}{30} = \frac{1}{12} \implies R_{\text{parallel1}} = 12 \, \Omega
\]
2. Calculate the equivalent resistance of \(R_4\) and \(R_5\) in parallel:
\[
\frac{1}{R_{\text{parallel2}}} = \frac{1}{R_4} + \frac{1}{R_5}
= \frac{1}{30} + \frac{1}{60} = \frac{1}{20} \implies R_{\text{parallel2}} = 20 \, \Omega
\]
3. Add \(R_3\) in series with the equivalent resistances from step 1 and step 2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1b551779-a94f-4658-8ab9-cb1269496a60%2F8a3853df-18d9-4eed-82ec-ba6493ec734e%2F89b202_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
For circuits in series and parallel, the supply voltage \(V_{\text{Supply}} = 240\) Volts. The resistor values are as follows:
- \(R_1 = 20 \, \Omega\)
- \(R_2 = 30 \, \Omega\)
- \(R_3 = 8 \, \Omega\)
- \(R_4 = 30 \, \Omega\)
- \(R_5 = 60 \, \Omega\)
What will be the total resistance value?
**Diagram Explanation:**
The diagram illustrates a circuit with a combination of series and parallel resistors:
- \(R_1\) and \(R_2\) are in parallel at the top part of the circuit.
- \(R_3\) is in series with the parallel combination of \(R_4\) and \(R_5\).
- \(R_4\) and \(R_5\) are in parallel at the bottom of the circuit.
**Answer Choices:**
- \(40 \, \Omega\)
- \(20 \, \Omega\)
- \(100 \, \Omega\)
- \(60 \, \Omega\)
**Solution Explanation:**
To find the total resistance:
1. Calculate the equivalent resistance of \(R_1\) and \(R_2\) in parallel:
\[
\frac{1}{R_{\text{parallel1}}} = \frac{1}{R_1} + \frac{1}{R_2}
= \frac{1}{20} + \frac{1}{30} = \frac{1}{12} \implies R_{\text{parallel1}} = 12 \, \Omega
\]
2. Calculate the equivalent resistance of \(R_4\) and \(R_5\) in parallel:
\[
\frac{1}{R_{\text{parallel2}}} = \frac{1}{R_4} + \frac{1}{R_5}
= \frac{1}{30} + \frac{1}{60} = \frac{1}{20} \implies R_{\text{parallel2}} = 20 \, \Omega
\]
3. Add \(R_3\) in series with the equivalent resistances from step 1 and step 2
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