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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 _

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Act like a helpful tutor and exlain me :Uploaded ImageTOTAL: 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 _

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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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