Question

For the given circuit, during maximum power transfer, the value of \(R\) and the magnitude of the power delivered to \(R\) are

• A. \(5~\Omega\) and \(1~W\)
• B. \(15~\Omega\) and \(10~W\)
• C. \(2.12~\Omega\) and \(1.233~W\)
• D. \(5.33~\Omega\) and \(0.188~W\)

Hint:

For maximum power transfer problems: \[ R_L=R_{th} \] and \[ P_{\max}=\frac{V_{th}^{2}}{4R_{th}}. \] Always find the Thevenin equivalent seen from the load terminals before applying the theorem.

Answer 1

The Correct Option is D

Concept: According to the Maximum Power Transfer Theorem, maximum power is delivered to a load when the load resistance equals the Thevenin resistance seen from the load terminals. Thus, \[ R=R_{th}. \] The maximum power delivered is \[ P_{\max} = \frac{V_{th}^{2}}{4R_{th}}. \] Therefore, we first determine the Thevenin equivalent resistance and Thevenin voltage across the terminals where \(R\) is connected.

Step 1: Find the Thevenin resistance seen by \(R\).
Remove the load resistance \(R\). To determine \(R_{th}\),
• Open the \(5~A\) current source.
• Short-circuit the \(24~V\) voltage source. The left terminal of \(R\) is then connected to ground through \[ 2\Omega. \] The right terminal of \(R\) is connected to ground through \[ 5\Omega \] in parallel with \[ 10\Omega. \] Hence, \[ R_{\text{right}} = \frac{5\times10}{5+10} = \frac{50}{15} = \frac{10}{3}\Omega. \] Since the two terminals are connected through ground, \[ R_{th} = 2+\frac{10}{3} = \frac{16}{3}\Omega. \] Therefore, \[ \boxed{R_{th}=5.33\Omega}. \] By maximum power transfer theorem, \[ \boxed{R=5.33\Omega}. \]

Step 2: Determine the Thevenin voltage \(V_{th}\).
With \(R\) removed, let the left terminal voltage be \(V_A\) and the right terminal voltage be \(V_B\). At node \(A\), the \(5A\) source injects current into the node and the only path to ground is through the \(2\Omega\) resistor. Hence, \[ V_A = 5\times2 = 10V. \] At node \(B\), the node is connected to \(24V\) through \(10\Omega\) and to ground through \(5\Omega\). Applying voltage division, \[ V_B = 24 \left( \frac{5}{10+5} \right). \] \[ = 8V. \] Thus, \[ V_{th} = V_A-V_B = 10-8 = 2V. \] \[ \boxed{V_{th}=2V}. \]

Step 3: Compute the maximum power delivered to \(R\).
Using \[ P_{\max} = \frac{V_{th}^{2}}{4R_{th}}, \] we get \[ P_{\max} = \frac{2^2}{4\times\frac{16}{3}}. \] \[ = \frac{4}{\frac{64}{3}}. \] \[ = \frac{12}{64}. \] \[ = 0.1875W. \] Therefore, \[ P_{\max} \approx 0.188W. \]

Step 4: Write the final answer.
Hence, \[ \boxed{ R=5.33\Omega \quad\text{and}\quad P_{\max}=0.188W } \] which corresponds to option (D).