Math Problem Statement

Calculate the current flowing out of the positive pole of each battery in the given circuit with three batteries (24V, 20V, 18V) and resistors (0.5Ω, 0.3Ω, 0.4Ω, 2Ω, 5Ω).

Solution

The problem asks us to calculate the current flowing out of the positive pole of each battery in the given circuit diagram. Let's walk through the solution step by step.


Step 1: Analyzing the circuit

The circuit shown has:

  1. Three batteries: 24V24 \, V, 20V20 \, V, and 18V18 \, V.
  2. Several resistors:
    • 2Ω2 \, \Omega and 5Ω5 \, \Omega in the upper branch.
    • 0.5Ω0.5 \, \Omega and 0.4Ω0.4 \, \Omega in the lower branches.
    • A shared 0.3Ω0.3 \, \Omega resistor in the middle branch.

Since the batteries have different voltages, we need to solve the circuit using Kirchhoff's laws (KVL and KCL) to determine the current in each branch.


Step 2: Setting up currents through each branch

We assume:

  • I1I_1: Current from the 24V24 \, V battery.
  • I2I_2: Current through the 20V20 \, V battery.
  • I3I_3: Current through the 18V18 \, V battery.

From the hint provided in the problem, it looks like:

  • I1=1.474AI_1 = 1.474 \, A
  • I2=1.04AI_2 = -1.04 \, A
  • I3=0.43AI_3 = -0.43 \, A

Step 3: Explanation of the current signs

The negative sign for the 20V20 \, V and 18V18 \, V batteries indicates that the assumed current direction was opposite to the actual direction.

  • 24V24 \, V battery: 1.474A1.474 \, A flows out of the positive terminal.
  • 20V20 \, V battery: 1.04A1.04 \, A flows into the positive terminal (since the actual current is negative).
  • 18V18 \, V battery: 0.43A0.43 \, A flows into the positive terminal (again, current is negative).

Step 4: Checking the validity using Kirchhoff’s Current Law (KCL)

The current through the shared resistor (0.3Ω0.3 \, \Omega) should satisfy the junction rule: I1=I2+I3I_1 = I_2 + I_3 Substitute the values: 1.474A=1.04A+(0.43A)1.474 \, A = -1.04 \, A + (-0.43 \, A) This equation holds, confirming the results.


Step 5: Results

  • Current out of 24 V battery: 1.474A1.474 \, A
  • Current out of 20 V battery: 1.04A-1.04 \, A (meaning 1.04A1.04 \, A flows into the battery)
  • Current out of 18 V battery: 0.43A-0.43 \, A (meaning 0.43A0.43 \, A flows into the battery)

Do you need further details or a more specific explanation?


5 Related Questions:

  1. What are Kirchhoff’s Voltage and Current Laws (KVL and KCL)?
  2. How do you analyze circuits with multiple batteries?
  3. Why are some currents negative, and what does that indicate about direction?
  4. How does resistance affect current in different branches of a circuit?
  5. What methods can be used to solve circuits other than Kirchhoff’s laws?

Tip: When dealing with multiple batteries, always assume initial current directions. If you get negative values, it just means the real direction is opposite!

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Math Problem Analysis

Mathematical Concepts

Kirchhoff's Laws
Current Calculation
Resistors in Series and Parallel

Formulas

Kirchhoff's Voltage Law (KVL): ΣV = 0 in a closed loop
Kirchhoff's Current Law (KCL): ΣI entering a junction = ΣI leaving the junction
Ohm's Law: V = IR

Theorems

Kirchhoff's Voltage Law
Kirchhoff's Current Law
Ohm's Law

Suitable Grade Level

Undergraduate Electrical Engineering