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)
Show Hint
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: Note what the ideal source does to the $5\ \Omega$ resistor.
The $10$ V source sits directly across the $5\ \Omega$ resistor, fixing that node at $10$ V regardless of current drawn. So whatever current the $5\ \Omega$ resistor pulls, it does not change the voltage feeding the rest of the network through the $10\ \Omega$ resistor. This means we can ignore the $5\ \Omega$ resistor for everything downstream of the source node. Step 2: Find the open-circuit voltage using a simple loop.
With $R_L$ disconnected, current only flows in the loop made of the $10$ V source, the $10\ \Omega$ resistor, and the $4\ \Omega$ resistor to ground. The loop current is
\[ I=\frac{10}{10+4}=0.7143\ \text{A} \]
The open-circuit voltage at the output port is the drop across the $4\ \Omega$ resistor:
\[ V_{th}=I\times4=0.7143\times4=2.857\ \text{V} \] Step 3: Find $R_{th}$ by injecting a test current.
Kill the $10$ V source (replace it with a short) and push a $1$ A test current into the output port. The $5\ \Omega$ resistor again carries no current since both its ends sit at ground once the source is shorted. The test current splits between the $10\ \Omega$ path (to the shorted source node, which is ground) and the $4\ \Omega$ path (straight to ground), both starting from the test-current node. Writing KCL there:
\[ 1=\frac{V_{test}}{10}+\frac{V_{test}}{4} \]
\[ 1=V_{test}(0.1+0.25)=0.35\,V_{test} \]
\[ V_{test}=\frac{1}{0.35}=2.857\ \text{V} \]
Since the test current was $1$ A,
\[ R_{th}=\frac{V_{test}}{I_{test}}=2.857\ \Omega \] Step 4: Set the load equal to $R_{th}$.
Maximum power reaches the load when $R_L=R_{th}$, so
\[ \boxed{R_L=2.86\ \Omega} \]
