A lossless transmission line having characteristic impedance \(Z_{0}=50 \Omega\) is driven by a source at the input end that consists of the series combination of a 10 -V sinusoidal generator and a \(50-\Omega\) resistor. The line is one- quarter wavelength long. At the other end of the line, a load impedance, \(Z_{L}=50-j 50 \Omega\) is attached. (a) Evaluate the input impedance to the line seen by the voltage source-resistor combination; \((b)\) evaluate the power that is dissipated by the load; \((c)\) evaluate the voltage amplitude that appears across the load.

First, we need to calculate the reflection coefficient (Gamma) at the load, which is given by: $$\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0}$$ Substitute the given values, \(Z_L=50-j50 \,\Omega\) and \(Z_0=50 \,\Omega\): $$\Gamma = \frac{(50-j 50) - 50}{(50-j 50) + 50}$$ $$\Gamma = -j\frac{50}{100} = -j0.5$$

Now, we need to find the input impedance \((Z_{in})\). As the line is one-quarter wavelength long, we can use the following formula: $$Z_{in} = Z_0 \frac{Z_L + j Z_0 \tan (\beta L)}{Z_0 + j Z_L \tan (\beta L)}$$ Since we are dealing with a lossless transmission line and considering that it is one-quarter wavelength long, we know that \(\beta L = \frac{\pi}{2}\) (90 degrees). Thus, the input impedance equation can be simplified to this: $$Z_{in} = Z_0 \frac{Z_L + j Z_0 }{Z_0 + j Z_L}$$ Substitute the given values: $$Z_{in} = 50 \frac{50-j 50 + j 50 }{50 + j (50-j 50)}$$ $$Z_{in} = 50 \Rightarrow$$ So, the input impedance seen by the voltage source-resistor combination is \(50 \, \Omega\).

To find the power dissipated by the load, we first need to determine the current in the circuit. Since the source has a series resistor of \(50 \, \Omega\), and we found the input impedance to be \(50 \, \Omega\), the total impedance is simply double the input impedance: $$Z_{total} = 2 Z_{in} = 2 (50) = 100 \, \Omega$$ Now, we can find the current from the voltage source and the total impedance: $$I = \frac{V}{Z_{total}} = \frac{10 \, V}{100 \, \Omega} = 0.1 \, A$$ The power dissipated by the load can be calculated using the following formula: $$P = \frac{1}{2}(I^2)R$$ Where \(R\) is the real part of the load impedance, which is \(50 \, \Omega\): $$P = \frac{1}{2}(0.1^2)(50) = 0.25 \, W$$ So, the power dissipated by the load is \(0.25 \, W\).

Lastly, we need to find the voltage amplitude across the load. We can use the Ohm's law to calculate the voltage: $$V_{L} = I \times Z_{L}$$ $$V_{L} = 0.1 \, A \times (50 - j50) \, \Omega$$ $$V_{L} = 5 - j5 \, V$$ So, the voltage amplitude across the load is \(5\sqrt{2} \, V\).