Calculate v and ix in the circuit

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Calculate v and ix in the circuit of Fig. 2.79.

The Correct Answer and Explanation is :

To solve for ( v ) and ( i_x ) in the given circuit, we will use Kirchhoff’s Voltage Law (KVL), Ohm’s Law, and the current divider rule. Here’s a step-by-step solution:

Analyzing the Circuit

  1. Identify Components and Values:
  • The circuit consists of three resistors (6 Ω, 3 Ω, and ( R )), and one independent current source of 3 A.
  • ( R ) has been replaced by a dependent voltage source ( 4i_x ).
  1. Labeling the Circuit:
  • ( i_1 ) is the current through the 6 Ω resistor.
  • ( i_2 ) is the current through the 3 Ω resistor.
  • ( i_x ) is the current through the dependent voltage source, also flowing through ( R ).
  1. Applying Kirchhoff’s Current Law (KCL):
  • At the node where the currents split, ( i_1 + i_2 = 3 ) A (current from the source).
  1. Applying Kirchhoff’s Voltage Law (KVL):
  • For the loop including the 6 Ω resistor and the dependent voltage source: ( 6i_1 + 4i_x = 0 ).
  • For the loop including all elements: ( 6i_1 + 3i_2 + 4i_x = 0 ).

Solving the Equations

  1. From KCL, ( i_1 + i_2 = 3 ).
  2. Using KVL, we can write ( 6i_1 + 4i_x = 0 ). Thus, ( i_x = -\frac{3}{2}i_1 ).
  3. Substitute ( i_x ) in the second KVL equation: ( 6i_1 + 3i_2 – 6i_1 = 0 ) simplifies to ( 3i_2 = 0 ), which implies ( i_2 = 0 ).
  4. Since ( i_1 + i_2 = 3 ), we get ( i_1 = 3 ) A.
  5. Now, substituting ( i_1 = 3 ) A back to find ( i_x ), ( i_x = -\frac{3}{2} \times 3 = -4.5 ) A.

Conclusion

  • ( i_x = -4.5 ) A: The negative sign indicates that the direction of ( i_x ) assumed in the analysis is opposite to the real direction.
  • ( v = 4i_x = 4 \times (-4.5) = -18 ) V: This is the voltage across the dependent source.

Explanation

In the given circuit, we used KCL to determine the distribution of the total current provided by the source among different branches. KVL helped in relating the voltage across different components in the loop. The negative sign on ( i_x ) is crucial as it indicates that the direction of the current through the dependent voltage source, as assumed, is opposite to the real current flow. The analysis shows that ( i_x ) is a significant current flowing in the opposite direction, which impacts the calculations for voltage ( v ). Such analysis is essential for designing and understanding complex circuits involving dependent sources.

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