Math Problem Statement

Solve the DC circuit with two resistors, R1 = 569 Ω and R2 = 287 Ω, and a total voltage of 74.31 V. Find the voltage, current, and power for each resistor.

Solution

The image you provided shows a DC circuit with two resistors, R1=569ΩR_1 = 569 \, \Omega and R2=287ΩR_2 = 287 \, \Omega, connected in series. The circuit has a voltage source of 74.31V74.31 \, V. We are tasked with calculating the voltage, current, and power for each resistor.

1. Total Resistance

Since the resistors are in series, the total resistance RtotalR_{\text{total}} is the sum of R1R_1 and R2R_2: Rtotal=R1+R2=569Ω+287Ω=856ΩR_{\text{total}} = R_1 + R_2 = 569 \, \Omega + 287 \, \Omega = 856 \, \Omega

2. Current in the Circuit

Using Ohm's law, the current II through the circuit can be found by dividing the total voltage by the total resistance: I=VtotalRtotal=74.31V856Ω0.0868A=86.8mAI = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{74.31 \, V}{856 \, \Omega} \approx 0.0868 \, A = 86.8 \, mA

3. Voltage Across Each Resistor

The voltage drop across each resistor can be found using Ohm's law (V=IRV = IR):

  • For R1R_1: V1=I×R1=0.0868A×569Ω49.42VV_1 = I \times R_1 = 0.0868 \, A \times 569 \, \Omega \approx 49.42 \, V

  • For R2R_2: V2=I×R2=0.0868A×287Ω24.89VV_2 = I \times R_2 = 0.0868 \, A \times 287 \, \Omega \approx 24.89 \, V

4. Power Dissipated by Each Resistor

The power dissipated by each resistor is given by P=I2×RP = I^2 \times R:

  • For R1R_1: P1=(0.0868A)2×569Ω4.27WP_1 = (0.0868 \, A)^2 \times 569 \, \Omega \approx 4.27 \, W

  • For R2R_2: P2=(0.0868A)2×287Ω2.16WP_2 = (0.0868 \, A)^2 \times 287 \, \Omega \approx 2.16 \, W

Summary of Results:

  • I=86.8mAI = 86.8 \, mA
  • Voltage across R1=49.42VR_1 = 49.42 \, V
  • Voltage across R2=24.89VR_2 = 24.89 \, V
  • Power dissipated by R1=4.27WR_1 = 4.27 \, W
  • Power dissipated by R2=2.16WR_2 = 2.16 \, W

Would you like further details on any of these steps or calculations?

Here are five related questions you might explore next:

  1. What is the total power dissipated in the circuit?
  2. How would the current change if R1R_1 were doubled?
  3. How would the voltage across R2R_2 change if the total voltage was reduced to 50V50 \, V?
  4. What would happen to the circuit if R2R_2 was removed (open circuit)?
  5. How does the power dissipation change if the resistors were arranged in parallel instead of series?

Tip: In a series circuit, the current remains the same through all components, but the voltage divides based on the resistance of each component.

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Math Problem Analysis

Mathematical Concepts

Ohm's Law
Series Circuit Analysis
Power Dissipation

Formulas

R_total = R1 + R2
I = V_total / R_total
V = I * R
P = I^2 * R

Theorems

Ohm's Law

Suitable Grade Level

Grades 9-12