Question Seven The sending end current of a transmission line is given by: \[ I_{s}=\frac{V_{s}}{Z_{o}} \text { tanh } P L \] Calculate the value of the sending current, in polar form, given: \( V_{s}=200 \mathrm{v}, Z_{o}=560+j 420 \Omega, P=0.20 \) and \( L=10 \)

To find the sending end current \( I_s \) in polar form, first, we need to calculate the characteristic impedance \( Z_o \) in polar form. 1. Find the magnitude and angle of \( Z_o \): \[ |Z_o| = \sqrt{(560)^2 + (420)^2} = \sqrt{313600 + 176400} = \sqrt{490000} = 700 \, \Omega \] \[ \theta = \tan^{-1}\left(\frac{420}{560}\right) = \tan^{-1}\left(\frac{3}{4}\right) \approx 36.87^\circ \] Thus, \( Z_o \) can be expressed as \( 700 \angle 36.87^\circ \Omega \). 2. Calculate \( \tanh(P L) \): \[ P L = 0.2 \times 10 = 2 \] Using a scientific calculator, \( \tanh(2) \approx 0.9640 \). 3. Substitute the values into the current equation: \[ I_s = \frac{V_s}{|Z_o|} \tanh(P L) = \frac{200 \, V}{700 \, \Omega} \times 0.9640 \approx \frac{192.8}{700} = 0.2754 \, A \] 4. In polar form: The angle is equal to the angle of \( Z_o \): \[ I_s \approx 0.2754 \angle 36.87^\circ \, A \] So, the sending current \( I_s \) in polar form is approximately \( 0.2754 \angle 36.87^\circ \, A \).