Math Problem Statement

Solve this combination circuit

Solve this combination circuit.

Solution

To solve the given DC circuit problem, we need to analyze the provided values and configurations of the resistors and voltages. Let's extract the relevant information from the image and work through the calculations:

Given Information:

  1. Total Voltage (Vt): 85.06 V85.06 \ \text{V}
  2. Total Resistance (Rt): 842.6 Ω842.6 \ \Omega
  3. Total Current (It): 100.95 mA100.95 \ \text{mA}
  4. Individual Resistances:
    • R1=450 ΩR_1 = 450 \ \Omega
    • R2=171 ΩR_2 = 171 \ \Omega
    • R3=398 ΩR_3 = 398 \ \Omega
    • R4=273 ΩR_4 = 273 \ \Omega
  5. Voltages and Currents through some resistors:
    • VR1=45.42 VV_{R_1} = 45.42 \ \text{V}, IR1=100.95 mAI_{R_1} = 100.95 \ \text{mA}
    • VR4=27.55 VV_{R_4} = 27.55 \ \text{V}, IR4=100.95 mAI_{R_4} = 100.95 \ \text{mA}

Circuit Analysis Approach:

  1. Identify Series and Parallel Connections:

    • R1R_1 and R4R_4 share the same current IR1=IR4I_{R_1} = I_{R_4}, indicating they are in series with the rest of the circuit.
    • R2R_2 and R3R_3 have a current of 0 mA, indicating an open circuit between them. This means no current flows through that part, and the series part with R1R_1 and R4R_4 dominates the behavior.
  2. Calculate Equivalent Resistance for Series Resistors: Since R1R_1 and R4R_4 are in series: Rseries=R1+R4=450 Ω+273 Ω=723 ΩR_{\text{series}} = R_1 + R_4 = 450 \ \Omega + 273 \ \Omega = 723 \ \Omega

  3. Verify Total Resistance Calculation: The circuit displays a total resistance of 842.6 Ω842.6 \ \Omega. With R2R_2 and R3R_3 having no current, it suggests they don't affect the total resistance directly. However, they might be in parallel with the series part but have an open connection, thus not contributing.

  4. Check Power Calculations: Power across each resistor can be verified using P=VIP = V \cdot I or P=I2RP = I^2 \cdot R:

    • Power across R1R_1: PR1=45.42 V×0.10095 A4.58 WP_{R_1} = 45.42 \ \text{V} \times 0.10095 \ \text{A} \approx 4.58 \ \text{W}
    • Power across R4R_4: PR4=27.55 V×0.10095 A2.78 WP_{R_4} = 27.55 \ \text{V} \times 0.10095 \ \text{A} \approx 2.78 \ \text{W}

Conclusion:

  • The primary active path involves R1R_1 and R4R_4 in series.
  • The open circuit between R2R_2 and R3R_3 results in no current through that branch.
  • The calculations for power and current match the given values for the resistors in the series circuit.

Would you like further details on the derivation or more explanations on series-parallel circuits? Here are 5 related questions that could deepen your understanding:

  1. How do you determine if resistors are in series or parallel in a complex circuit?
  2. What happens to the total current in a parallel circuit when one branch is open?
  3. How can you calculate power dissipation across multiple resistors?
  4. What is the significance of Kirchhoff’s Voltage Law (KVL) in analyzing circuits like this?
  5. How do changes in resistance affect the overall current in a circuit?

Tip: In series circuits, the same current flows through each component, while in parallel circuits, the voltage across each component is the same.

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Math Problem Analysis

Mathematical Concepts

Ohm's Law
Series and Parallel Circuits
Power Dissipation

Formulas

V = IR (Ohm's Law)
P = IV (Power calculation)
Series resistance: Rs = R1 + R2 + ...
Parallel resistance: 1/Rp = 1/R1 + 1/R2 + ...

Theorems

Kirchhoff's Voltage Law (KVL)
Kirchhoff's Current Law (KCL)

Suitable Grade Level

Grades 10-12 (High School)