Answer:
Explanation:
Parameters
RL = load resistance
PL = Power output to the load
IL = load current
Rs = internal resistance of the source
Is = source current
in the first condition
RL = 50 ohms, PL = 0.68055555555555
in the second case
RL = 460 ohms PL = 1.11783376314223
the diagram is as shown attached below
[tex]I_l = I_s * \frac{R_s}{R_s + R_l}............1\\\\P_l = (I_l)^{2} * R_l ...................2\\\\[/tex]
Substituting for Il in eqn 2
[tex]P_l = (I_s)^{2} * \frac{R_s^{2} * R_l}{(R_s + R_l)^{2} } ............3[/tex]
Using equation 3 for both cases
case 1
[tex]0.68055555555555 = (I_s)^{2} * \frac{R_s^{2} * 50}{(R_s + 50)^{2} } ............4[/tex]
case 2
[tex]1.11783376314223 = (I_s)^{2} * \frac{R_s^{2} * 460}{(R_s + 460)^{2} } ............5[/tex]
solving equation 4 and 5 we get
[tex]I_s = 0.07063 A\\\\R_s = 39.12[/tex]
[a]
when Vl * il is maximum then
Rl = Rs
⇒ Rl = 39.12 Ohms
Using equation 1
[tex]I_l = I_s * \frac{R_s}{R_s + R_l}............1\\\\I_l = 0.07064 * \frac{39.12}{39.12 + 39.12}\\\\I_l = 0.0353[/tex]
Vl = il * Rl = 0.0353 * 39.12 = 1.3817 V
hence,
IL = 0.0353 A RL = 39.12 Ohms VL= 1.38 V
[b]
When VL is maximum
VL is max when RL = ∞ (infinite)
from equation 1
IL = 0
VL = Is * Rs = 0.0764 * 39.12 = 2.763 V
[c]
when IL is a Maximum
⇒ RL = 0 ohms
substituting RL = 0 in equation 1
IL = Is = 0.0764 A
VL = IL * RL = 0.0764 * 0 = 0
