Math Problem Statement

In the figure, the current in resistance R6 is i6 = 1.31 A and the resistances are R1 = R2 = R3 = 1.68Ω, R4 = 17.7Ω, R5 = 7.71Ω, and R6 = 4.65Ω. What is the emf of the ideal battery?

Solution

To find the electromotive force (emf) of the ideal battery in this circuit, we can use Ohm's Law and Kirchhoff's Loop Rule. Let's break it down step by step:

Circuit Configuration:

  • The circuit has 6 resistors: R1R_1, R2R_2, R3R_3, R4R_4, R5R_5, and R6R_6.
  • The current i6=1.31Ai_6 = 1.31 \, \text{A} flows through R6=4.65ΩR_6 = 4.65 \, \Omega.

Given Values:

  • R1=R2=R3=1.68ΩR_1 = R_2 = R_3 = 1.68 \, \Omega
  • R4=17.7ΩR_4 = 17.7 \, \Omega
  • R5=7.71ΩR_5 = 7.71 \, \Omega
  • R6=4.65ΩR_6 = 4.65 \, \Omega
  • i6=1.31Ai_6 = 1.31 \, \text{A}

Step 1: Calculate Voltage Drop across R6R_6

The voltage across R6R_6 is given by Ohm's Law: V6=i6R6=1.31A×4.65Ω=6.0915VV_6 = i_6 \cdot R_6 = 1.31 \, \text{A} \times 4.65 \, \Omega = 6.0915 \, \text{V}

Step 2: Simplify the Circuit

  • R1R_1, R2R_2, and R3R_3 are in series, so their equivalent resistance is: R123=R1+R2+R3=1.68Ω+1.68Ω+1.68Ω=5.04ΩR_{123} = R_1 + R_2 + R_3 = 1.68 \, \Omega + 1.68 \, \Omega + 1.68 \, \Omega = 5.04 \, \Omega
  • R4R_4, R5R_5, and R6R_6 are in series, so their equivalent resistance is: R456=R4+R5+R6=17.7Ω+7.71Ω+4.65Ω=30.06ΩR_{456} = R_4 + R_5 + R_6 = 17.7 \, \Omega + 7.71 \, \Omega + 4.65 \, \Omega = 30.06 \, \Omega

Step 3: Total Resistance of the Circuit

Since the two series combinations are in parallel, the total resistance RtotalR_{\text{total}} is: 1Rtotal=1R123+1R456=15.04+130.06\frac{1}{R_{\text{total}}} = \frac{1}{R_{123}} + \frac{1}{R_{456}} = \frac{1}{5.04} + \frac{1}{30.06} Solving this: 1Rtotal0.1984+0.0333=0.2317\frac{1}{R_{\text{total}}} \approx 0.1984 + 0.0333 = 0.2317 Rtotal4.32ΩR_{\text{total}} \approx 4.32 \, \Omega

Step 4: Apply Ohm's Law to Find EMF

Using Ohm's Law for the whole circuit, we can calculate the emf (E\mathcal{E}) by multiplying the total resistance by the current i6i_6, which is the same through the entire circuit: E=i6Rtotal=1.31A×4.32Ω=5.66V\mathcal{E} = i_6 \cdot R_{\text{total}} = 1.31 \, \text{A} \times 4.32 \, \Omega = 5.66 \, \text{V}

Thus, the emf of the ideal battery is approximately 5.66 V.

Would you like further clarification or details?

Related Questions:

  1. How does changing R6R_6 affect the total resistance of the circuit?
  2. What would happen if R4R_4 were removed from the circuit?
  3. How would the total current change if R5R_5 were doubled?
  4. What is the significance of the parallel configuration in determining the total resistance?
  5. Can you calculate the power dissipated in R6R_6?

Tip:

When analyzing complex circuits, simplifying series and parallel resistances step by step can help make the analysis much easier!

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Math Problem Analysis

Mathematical Concepts

Ohm's Law
Kirchhoff's Loop Rule
Series and Parallel Circuits

Formulas

Ohm's Law: V = IR
Series Resistance: R_total = R1 + R2 + ...
Parallel Resistance: 1/R_total = 1/R1 + 1/R2 + ...

Theorems

Kirchhoff's Circuit Laws

Suitable Grade Level

Grades 9-12