Let's break the problem into parts and solve step by step. ### **Given Data:** - **Generator open-circuit voltage:** \( v_g(t) = 10 \sin(8000\pi t)\ \mathrm{V} \) - **Generator internal impedance:** \( Z_g = 40 + j30\ \Omega \) - **Transmission line characteristic impedance:** \( Z_0 = 50\ \Omega \) (matched) - **Line resistance per meter:** \( R' = 0.5\ \Omega/\mathrm{m} \) - **Loss tangent:** \( \tan\delta = 0.18\% = 0.0018 \) - **Line length:** \( l = 50\ \mathrm{m} \) - **Matched load:** \( Z_L = Z_0 = 50\ \Omega \) ### **Step 1: Angular frequency** From the source: \[ v_g(t) = 10 \sin(8000\pi t) \] So, \[ \omega = 8000\pi\ \mathrm{rad/s} \] \[ f = \frac{\omega}{2\pi} = 4000\ \mathrm{Hz} \] --- ## **Step 2: Line parameters** Assuming a low-loss or distortionless line (as stated), and given loss tangent: - **Loss tangent:** \(\tan\delta = \frac{\sigma}{\omega \epsilon} = 0.0018\) We'll need to find the propagation constant \(\gamma\) and characteristic impedance \(Z_0\). ### **Characteristic Impedance \(Z_0\)** Given as \(50~\Omega\), so we do **not** need to calculate. --- ### **Propagation Constant \(\gamma\) and Attenuation** For a distortionless line: \[ \gamma = \alpha + j\beta \] Where \(\alpha\) is the attenuation constant (Np/m), and \(\beta\) is the phase constant (rad/m). For a distortionless line: \[ \frac{R'}{L'} = \frac{G'}{C'} \] But we have a small loss tangent, so assume a low-loss line. #### **Attenuation constant:** \[ \alpha \approx \frac{R'}{2Z_0} \] Given: \[ R' = 0.5~\Omega/\mathrm{m},\qquad Z_0 = 50~\Omega \] \[ \alpha = \frac{0.5}{2 \times 50} = \frac{0.5}{100} = 0.005~\mathrm{Np/m} \] #### **Phase constant:** For low-loss lines, \(\beta \approx \omega \sqrt{L'C'}\), but since the line is distortionless and matched, and we have no velocity given, let's just use \[ v_p \approx c = 3 \times 10^8~\mathrm{m/s} \] \[ \beta = \frac{\omega}{v_p} = \frac{8000\pi}{3 \times 10^8} \approx 8.3776 \times 10^{-5}~\mathrm{rad/m} \] --- ## **Step 3: Input voltage to the line** The generator is connected to the line via its internal impedance, so the voltage **at the input of the line** is: Let: - \( V_g = 10\angle{-90^\circ}~\mathrm{V} \) (since \( \sin(\omega t) = \Im\{e^{j\omega t}\} \), or equivalently \( 10\sin(\omega t) = 10\cos(\omega t - 90^\circ) \)). The line is matched (\(Z_0 = Z_L = 50~\Omega\)), so no reflection. The voltage divider at the input: \[ V_{in} = V_g \cdot \frac{Z_0}{Z_g + Z_0} \] \[ Z_g + Z_0 = (40 + j30) + 50 = (90 + j30)~\Omega \] \[ |Z_g + Z_0| = \sqrt{90^2 + 30^2} = \sqrt{8100 + 900} = \sqrt{9000} = 94.87~\Omega \] \[ \angle Z_g + Z_0 = \tan^{-1} \left( \frac{30}{90} \right) = \tan^{-1}(1/3) \approx 18.43^\circ \] \[ V_{in} = 10\angle{-90^\circ} \cdot \frac{50}{94.87} \angle{-18.43^\circ} = 10 \cdot 0.527 \angle{-108.43^\circ} = 5.27 \angle{-108.43^\circ}~\mathrm{V} \] --- ## **Step 4: Forward voltage and current on the line** For a matched line (no reflection): The forward voltage at the input (\(z=0\)): \[ \tilde{V}_+(0) = V_{in} \] The forward current: \[ \tilde{I}_+(0) = \frac{\tilde{V}_+(0)}{Z_0} = \frac{5.27 \angle{-108.43^\circ}}{50} = 0.1054 \angle{-108.43^\circ}~\mathrm{A} \] The voltage and current at any point \(z\) along the line (\(z=0\) at the generator, \(z=l\) at the load): \[ \tilde{V}(z) = \tilde{V}_+(0) e^{-\gamma z} \] \[ \tilde{I}(z) = \frac{\tilde{V}_+(0)}{Z_0} e^{-\gamma z} \] Where: \[ \gamma = \alpha + j\beta \] \[ e^{-\gamma z} = e^{-\alpha z} e^{-j\beta z} \] --- ## **Step 5: Instantaneous expressions at arbitrary location (\(z\)), and at the load (\(z=l\))** ### **(a) At arbitrary location \(z\):** \[ v(z, t) = |\tilde{V}_+(0)| e^{-\alpha z} \sin \left( \omega t + \phi_{V_+} - \beta z \right ) \] where \(\phi_{V_+} = -108.43^\circ \) \[ i(z, t) = \frac{|\tilde{V}_+(0)|}{Z_0} e^{-\alpha z} \sin \left( \omega t + \phi_{V_+} - \beta z \right ) \] **Plug in numbers:** - \( |\tilde{V}_+(0)| = 5.27~\mathrm{V} \) - \( \phi_{V_+} = -108.43^\circ = -1.893~\mathrm{rad} \) - \( \alpha = 0.005~\mathrm{Np/m} \) - \( \beta = 8.3776 \times 10^{-5}~\mathrm{rad/m} \) - \( Z_0 = 50~\Omega \) - \( \omega = 8000\pi~\mathrm{rad/s} \) Thus, \[ \boxed{ v(z, t) = 5.27\, e^{-0.005 z} \sin \left( 8000\pi t - 1.893 - 8.38 \times 10^{-5} z \right )~\mathrm{V} } \] \[ \boxed{ i(z, t) = 0.1054\, e^{-0.005 z} \sin \left( 8000\pi t - 1.893 - 8.38 \times 10^{-5} z \right )~\mathrm{A} } \] --- ### **(b) At the load (\(z = l = 50~\mathrm{m}\))** First, calculate the attenuation factor: \[ e^{-\alpha l} = e^{-0.005 \times 50} = e^{-0.25} \approx 0.7788 \] \[ \beta l = 8.38 \times 10^{-5} \times 50 = 0.00419~\mathrm{rad} \] So, at the load: \[ v(l, t) = 5.27 \times 0.7788 \sin \left( 8000\pi t - 1.893 - 0.00419 \right ) \] \[ \approx 4.10 \sin (8000\pi t - 1.897 )~\mathrm{V} \] \[ i(l, t) = 0.1054 \times 0.7788 \sin \left( 8000\pi t - 1.897 \right ) \] \[ \approx 0.0821 \sin (8000\pi t - 1.897 )~\mathrm{A} \] \[ \boxed{ v(l, t) = 4.10 \sin (8000\pi t - 1.897)~\mathrm{V} } \] \[ \boxed{ i(l, t) = 0.0821 \sin (8000\pi t - 1.897)~\mathrm{A} } \] --- ### **(c) Average power delivered to the load** For a matched, lossy line, the average power delivered to the load is: \[ P_{avg} = \frac{1}{2} \frac{|V_{load}|^2}{Z_0} \] Where \( V_{load} \) is the **phasor amplitude** at the load (\(z = l\)): \( |V_{load}| = 4.10~\mathrm{V} \) \[ P_{avg} = \frac{1}{2} \frac{(4.10)^2}{50} = \frac{1}{2} \cdot \frac{16.81}{50} = \frac{8.405}{50} = 0.168~\mathrm{W} \] --- ## **Final answers (boxed):** --- ### **(a) Voltage and current at arbitrary location \(z\) along the line:** \[ \boxed{ \begin{aligned} v(z,t) &= 5.27\,e^{-0.005 z} \sin(8000\pi t - 1.893 - 8.38\times 10^{-5} z)\ \mathrm{V} \\ i(z,t) &= 0.1054\,e^{-0.005 z} \sin(8000\pi t - 1.893 - 8.38\times 10^{-5} z)\ \mathrm{A} \end{aligned} } \] --- ### **(b) Voltage and current at the load (\(z=l=50~\mathrm{m}\)):** \[ \boxed{ \begin{aligned} v(l, t) &= 4.10\,\sin(8000\pi t - 1.897)\ \mathrm{V} \\ i(l, t) &= 0.0821\,\sin(8000\pi t - 1.897)\ \mathrm{A} \end{aligned} } \] --- ### **(c) Average power delivered to the load:** \[ \boxed{ P_{avg} = 0.168~\mathrm{W} } \] --- #### **Summary Table** | Quantity | Expression | |----------|------------| | \(v(z, t)\) | \(5.27\,e^{-0.005 z} \sin(8000\pi t - 1.893 - 8.38\times 10^{-5} z)\ \mathrm{V}\) | | \(i(z, t)\) | \(0.1054\,e^{-0.005 z} \sin(8000\pi t - 1.893 - 8.38\times 10^{-5} z)\ \mathrm{A}\) | | \(v(l, t)\) | \(4.10\,\sin(8000\pi t - 1.897)\ \mathrm{V}\) | | \(i(l, t)\) | \(0.0821\,\sin(8000\pi t - 1.897)\ \mathrm{A}\) | | \(P_{avg}\) | \(0.168~\mathrm{W}\) | --- **If you need clarification on any step, or want to see phasor diagrams or waveforms, let me know!**