Math Problem Statement

solve using nodal

Using Nodal Analysis, solve for the current I3 in the figure below.

Solution

To solve for the current I3I_3 using Nodal Analysis, let us systematically proceed as follows.


Step 1: Define Node Voltages

Let:

  • VNV_N = voltage at node NN (common node between R1R_1, R2R_2, and R3R_3).

Step 2: Write KCL at Node NN

By Kirchhoff's Current Law (KCL), the sum of currents entering or leaving the node is zero. Assume the currents flow out of node NN:

VNV1R1+VNV2R2+VNVGR3=0\frac{V_N - V_1}{R_1} + \frac{V_N - V_2}{R_2} + \frac{V_N - V_G}{R_3} = 0

  • V1=30VV_1 = 30 \, \text{V}
  • V2=10VV_2 = 10 \, \text{V}
  • VG=0V (ground reference)V_G = 0 \, \text{V (ground reference)}
  • R1=1.2kΩ=1200ΩR_1 = 1.2 \, \text{k}\Omega = 1200 \, \Omega
  • R2=800ΩR_2 = 800 \, \Omega
  • R3=120ΩR_3 = 120 \, \Omega

Substitute the values into the KCL equation:

VN301200+VN10800+VN0120=0\frac{V_N - 30}{1200} + \frac{V_N - 10}{800} + \frac{V_N - 0}{120} = 0


Step 3: Simplify the Equation

Combine the terms:

VN301200+VN10800+VN120=0\frac{V_N - 30}{1200} + \frac{V_N - 10}{800} + \frac{V_N}{120} = 0

To eliminate fractions, find the LCM of the denominators 1200,800,1201200, 800, 120, which is 24002400. Multiply through by 24002400:

2400VN301200+2400VN10800+2400VN120=02400 \cdot \frac{V_N - 30}{1200} + 2400 \cdot \frac{V_N - 10}{800} + 2400 \cdot \frac{V_N}{120} = 0

Simplify each term:
2(VN30)+3(VN10)+20VN=02(V_N - 30) + 3(V_N - 10) + 20V_N = 0

Expand:
2VN60+3VN30+20VN=02V_N - 60 + 3V_N - 30 + 20V_N = 0

Combine like terms:
(2+3+20)VN90=0(2 + 3 + 20)V_N - 90 = 0

25VN=9025V_N = 90

Solve for VNV_N:
VN=9025=3.6VV_N = \frac{90}{25} = 3.6 \, \text{V}


Step 4: Solve for I3I_3

The current I3I_3 flows through resistor R3R_3, and it is given by Ohm's Law:

I3=VNVGR3I_3 = \frac{V_N - V_G}{R_3}

Substitute VN=3.6VV_N = 3.6 \, \text{V}, VG=0VV_G = 0 \, \text{V}, and R3=120ΩR_3 = 120 \, \Omega:

I3=3.60120=3.6120=0.03AI_3 = \frac{3.6 - 0}{120} = \frac{3.6}{120} = 0.03 \, \text{A}

Convert to milliamps (mA):

I3=30mAI_3 = 30 \, \text{mA}


Final Answer:

The current I3I_3 is 30 mA.


Do you need a detailed breakdown of any step? Let me know!

Here are related questions for practice:

  1. What is the current through R1R_1 and R2R_2?
  2. Verify the voltage at node NN using KCL.
  3. How would the current I3I_3 change if R3R_3 is doubled?
  4. Solve for the power dissipated in each resistor.
  5. How would the solution change if V2=20VV_2 = 20 \, \text{V}?

Tip: Always double-check units when solving for current and voltage in circuit analysis!

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Math Problem Analysis

Mathematical Concepts

Nodal Analysis
Kirchhoff's Current Law (KCL)
Ohm's Law

Formulas

Kirchhoff's Current Law: ΣI = 0
Ohm's Law: I = V / R

Theorems

Kirchhoff's Current Law
Ohm's Law

Suitable Grade Level

Undergraduate Engineering or Grades 11-12 Physics