Consider the two-port network shown. For maximum power transfer to the resistive load (\(R_L\)), the value of \(R_L\) should be \(\Omega\)
(Round off to two decimal places)
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Maximum power transfer needs the load resistance to equal the Thevenin resistance seen at the load terminals; deactivate the source and combine the remaining resistors.
Step 1: Recall the maximum power transfer condition.
A resistive load draws the maximum possible power from a network when its resistance equals the Thevenin resistance of the network as seen from the load terminals. So the target is to find \(R_{th}\) looking into the port where \(R_L\) sits. Step 2: Reduce the network for the open-circuit voltage.
The \(10\ \text{V}\) ideal source is applied directly across the \(5\ \Omega\) resistor, so that node is fixed at \(10\ \text{V}\) no matter what current the \(5\ \Omega\) resistor draws. This means the \(5\ \Omega\) resistor plays no role in what reaches the output port, since the source voltage feeding the rest of the circuit is already fixed.
With \(R_L\) removed, the only path left to the output node is the \(10\ \Omega\) resistor from the \(10\ \text{V}\) node, followed by the \(4\ \Omega\) resistor to ground. This is a simple voltage divider:
\[
V_{th}=10\times\frac{4}{10+4}=10\times\frac{4}{14}=\frac{40}{14}=2.857\ \text{V}
\] Step 3: Find the Thevenin resistance.
To get \(R_{th}\), deactivate the independent source by replacing it with a short circuit. Once the \(10\ \text{V}\) source is shorted, one end of the \(5\ \Omega\) resistor is grounded and the other end is also grounded through the short, so no current can ever flow in it, and it drops out entirely.
What remains, looking into the output port, is the \(10\ \Omega\) resistor (now going from the output node to ground through the shorted source) in parallel with the \(4\ \Omega\) resistor (already going from the output node to ground):
\[
R_{th}=\frac{10\times4}{10+4}=\frac{40}{14}=2.857\ \Omega
\] Step 4: Apply the maximum power transfer condition.
For maximum power transfer,
\[
R_L=R_{th}
\] Final Answer:
Rounding to two decimal places,
\[
\boxed{R_L=2.86\ \Omega}
\]
