Math Problem Statement

In the circuit of Fig. 3-39, calculate I and also the power absorbed by the dependent source. The answer is I = 2 A, and the power absorbed is 560 W.

Solution

In the given circuit (Fig. 3-39), we are tasked with calculating the current II and the power absorbed by the dependent source. Here’s a breakdown of the key steps to solve this:

Step 1: Analyzing the Circuit Components

  • There is a 16 A current source connected in parallel with:
    • A 14 Ω resistor,
    • A current-dependent current source (which depends on the current II through the 70 Ω resistor),
    • A 70 Ω resistor where the current is II.

Step 2: Apply Kirchhoff's Current Law (KCL) at the Node

Since the 16 A current splits between the 14 Ω resistor, the dependent current source (with a gain of 2), and the 70 Ω resistor, KCL at the node gives: 16=V14+2I+I16 = \frac{V}{14} + 2I + I Where:

  • V14\frac{V}{14} is the current through the 14 Ω resistor,
  • II is the current through the 70 Ω resistor, and
  • 2I2I is the current provided by the dependent source.

Step 3: Express the Voltage in Terms of II

The voltage across the 70 Ω resistor is: V=I×70V = I \times 70 Substitute this into the equation for the current through the 14 Ω resistor: 16=I×7014+2I+I16 = \frac{I \times 70}{14} + 2I + I Simplify: 16=5I+2I+I16 = 5I + 2I + I 16=8I16 = 8I Solving for II: I=168=2 AI = \frac{16}{8} = 2 \text{ A}

Step 4: Calculate Power Absorbed by the Dependent Source

The dependent current source provides 2I=2×2=4 A2I = 2 \times 2 = 4 \text{ A}. The voltage across it is the same as across the 70 Ω resistor, which is V=I×70=2×70=140 VV = I \times 70 = 2 \times 70 = 140 \text{ V}.

Thus, the power absorbed by the dependent source is: P=V×Idependent=140×4=560 WP = V \times I_{\text{dependent}} = 140 \times 4 = 560 \text{ W}

Final Answers:

  • The current I=2 AI = 2 \text{ A},
  • The power absorbed by the dependent source is 560 W560 \text{ W}.

Would you like more details on any of these steps?

Here are some related questions you can explore:

  1. How would the circuit change if the dependent source was voltage-dependent?
  2. What would happen if the 70 Ω resistor was removed from the circuit?
  3. How do dependent current sources behave in different circuit configurations?
  4. How is power dissipation calculated for passive and active circuit elements?
  5. How does Kirchhoff's Current Law apply in complex circuits with multiple nodes?

Tip: When solving circuit problems, always verify if the dependent source is current- or voltage-dependent to apply the correct relationships!

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Math Problem Analysis

Mathematical Concepts

Circuit Analysis
Ohm's Law
Kirchhoff's Current Law (KCL)
Power Calculation

Formulas

Kirchhoff's Current Law (KCL): ΣI = 0 at a node
Ohm's Law: V = IR
Power: P = VI

Theorems

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

Suitable Grade Level

Undergraduate Electrical Engineering