Math Problem Statement

Using superposition, calculate the current I labeled in the circuit below (downward through R2), giving your answer in A, to one decimal place. The labeled quantities are as follows: Vs = 3 V, Is = 2 A, R1 = 2 Ω, R2 = 3 Ω.

Solution

We are tasked with calculating the current labeled II using the superposition theorem for the given circuit. The labeled quantities are:

  • Vs=3VV_s = 3 \, \text{V}
  • Is=2AI_s = 2 \, \text{A}
  • R1=2ΩR_1 = 2 \, \Omega
  • R2=3ΩR_2 = 3 \, \Omega

Applying Superposition Theorem

The superposition theorem states that in a linear circuit with multiple independent sources, the total current or voltage for any element can be found by summing the individual contributions from each independent source, with all other sources replaced by their internal resistances (i.e., voltage sources are short-circuited and current sources are open-circuited).

Step 1: Contribution of the Voltage Source (VsV_s)

  • Set the current source IsI_s to 0 A (open circuit).
  • The circuit becomes a simple series circuit with VsV_s, R1R_1, and R2R_2.

The total resistance in the series is: Rtotal=R1+R2=2Ω+3Ω=5ΩR_{\text{total}} = R_1 + R_2 = 2 \, \Omega + 3 \, \Omega = 5 \, \Omega

Using Ohm's Law, the current through the series is: IVs=VsRtotal=3V5Ω=0.6AI_{\text{Vs}} = \frac{V_s}{R_{\text{total}}} = \frac{3 \, \text{V}}{5 \, \Omega} = 0.6 \, \text{A}

Since R1R_1 and R2R_2 are in series, this is the current through both resistors. The current through R2R_2 due to VsV_s is: IR2(fromVs)=0.6A(downward throughR2)I_{R_2}(\text{from} \, V_s) = 0.6 \, \text{A} \, (\text{downward through} \, R_2)

Step 2: Contribution of the Current Source (IsI_s)

  • Set the voltage source VsV_s to 0 V (short circuit).
  • The circuit now consists of R1R_1 and R2R_2 in parallel with Is=2AI_s = 2 \, \text{A}.

The equivalent resistance of R1R_1 and R2R_2 in parallel is: 1Req=1R1+1R2=12Ω+13Ω=56Ω\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2 \, \Omega} + \frac{1}{3 \, \Omega} = \frac{5}{6} \, \Omega Thus, Req=65Ω=1.2ΩR_{\text{eq}} = \frac{6}{5} \, \Omega = 1.2 \, \Omega

The current divides between R1R_1 and R2R_2. Using the current division rule, the current through R2R_2 is: IR2(fromIs)=Is×R1R1+R2=2A×2Ω2Ω+3Ω=2A×25=0.8AI_{R_2}(\text{from} \, I_s) = I_s \times \frac{R_1}{R_1 + R_2} = 2 \, \text{A} \times \frac{2 \, \Omega}{2 \, \Omega + 3 \, \Omega} = 2 \, \text{A} \times \frac{2}{5} = 0.8 \, \text{A}

This current is also downward through R2R_2.

Step 3: Total Current through R2R_2

Now, we sum the contributions from both sources: IR2=IR2(fromVs)+IR2(fromIs)=0.6A+0.8A=1.4AI_{R_2} = I_{R_2}(\text{from} \, V_s) + I_{R_2}(\text{from} \, I_s) = 0.6 \, \text{A} + 0.8 \, \text{A} = 1.4 \, \text{A}

Final Answer:

The current through R2R_2 (downward) is I=1.4AI = 1.4 \, \text{A}.


Would you like more details on any of the steps?

Here are 5 related questions you might find interesting:

  1. How would the result change if R1R_1 or R2R_2 were different values?
  2. How do you calculate current division in a parallel circuit?
  3. What would happen if VsV_s or IsI_s were reversed in polarity?
  4. How does superposition apply to circuits with dependent sources?
  5. Can superposition be used for circuits with nonlinear components?

Tip: Always remember to turn off all but one independent source at a time when using superposition.

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Math Problem Analysis

Mathematical Concepts

Superposition Theorem
Ohm's Law
Current Division
Series and Parallel Circuits

Formulas

Ohm's Law: I = V / R
Total resistance in series: R_total = R1 + R2
Equivalent resistance in parallel: 1 / Req = 1 / R1 + 1 / R2
Current division rule: I_R = I_total * (R_other / (R1 + R2))

Theorems

Superposition Theorem
Ohm's Law

Suitable Grade Level

Undergraduate Electrical Engineering