Act like a helpful tutor and exlain me :TOTAL: 30 POINTS 1- (10 points) In the following circuit find the current ia and the voltage across R4 #0 a 20 R2 M Vs RI < R4 Vv. 24V R3 200Q R4 ® 50Q _
Question:
Act like a helpful tutor and exlain me :
TOTAL: 30 POINTS
1- (10 points) In the following circuit find the current ia and the voltage across R4
#0 a 20
R2 M
Vs RI < R4 Vv.
24V R3 200Q R4
® 50Q _
This is AI generated Response.
Asked by: Alok Kumar
Created at: 2026-01-27 09:25:25
Asked by: Alok Kumar
Created at: 2026-01-27 09:25:25
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Let's break this problem down step by step!
## **Given:**
- Voltage source \(V_S = 24\,V\)
- \(R_1 = 48\,\Omega\)
- \(R_2 = 32\,\Omega\)
- \(R_3 = 50\,\Omega\)
- \(R_4 = 200\,\Omega\)
## **What to Find:**
1. The current \(i_a\) (total current from the source)
2. The voltage across \(R_4\) (\(V_{R4}\))
---
## **Step 1: Simplify the Circuit**
### **Identify Series and Parallel Combinations**
- \(R_1\) and \(R_2\) are in series.
- \(R_3\) and \(R_4\) are in parallel **with each other**, and this parallel group is in series with \(R_1\) and \(R_2\).
### **Find Equivalent Resistance of \(R_3\) and \(R_4\) in Parallel:**
\[
R_{34} = \left(\frac{1}{R_3} + \frac{1}{R_4}\right)^{-1} = \left(\frac{1}{50} + \frac{1}{200}\right)^{-1}
\]
\[
\frac{1}{R_{34}} = \frac{1}{50} + \frac{1}{200} = \frac{4}{200} + \frac{1}{200} = \frac{5}{200} = \frac{1}{40}
\]
\[
R_{34} = 40\,\Omega
\]
### **Find Total Resistance in the Circuit:**
\[
R_{\text{total}} = R_1 + R_2 + R_{34}
\]
\[
= 48 + 32 + 40 = 120\,\Omega
\]
---
## **Step 2: Find the Current \(i_a\)**
Using Ohm's Law (\(V = IR\)):
\[
i_a = \frac{V_S}{R_{\text{total}}} = \frac{24}{120} = .2\,A
\]
---
## **Step 3: Find the Voltage Across \(R_4\) (\(V_{R4}\))**
First, find the voltage across the parallel combination (\(R_3\) and \(R_4\)).
The voltage across the parallel branch, \(V_{34}\):
\[
V_{34} = \text{Current through series resistors} \times R_{34}
\]
But it's easier to use:
\[
V_{34} = V_S - (\text{Voltage drops across } R_1 \text{ and } R_2)
\]
#### Calculate the voltage drop across \(R_1\) and \(R_2\):
\[
V_{R_1} = i_a \times R_1 = .2 \times 48 = 9.6\,V
\]
\[
V_{R_2} = i_a \times R_2 = .2 \times 32 = 6.4\,V
\]
\[
V_{34} = V_S - (V_{R_1} + V_{R_2}) = 24 - (9.6 + 6.4) = 24 - 16 = 8\,V
\]
#### So, the voltage across \(R_4\) is:
\[
V_{R4} = V_{34} = 8\,V
\]
---
## **Step 4: Final Answers**
**1. The current \(i_a\) is:**
\[
\boxed{.2\,A}
\]
**2. The voltage across \(R_4\) (\(V_{R4}\)) is:**
\[
\boxed{8\,V}
\]
---
### **Summary Table**
| Quantity | Value |
|--------------|----------|
| \(i_a\) | .2 A |
| \(V_{R4}\) | 8 V |
---
**If you need help finding any currents through specific resistors, let me know!**
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