A certain practical current source provides 0.680555555555555 W to a 50 Ω load. That same practical current source provides 1.11783376314223 W to an 460 Ω load. If the same practical current source had a resistance RL is connected to it, creating a voltage across RL of vL and a current through RL of iL. Find the values of RL, vL, and iL if
(a) vL×iL is maximum
(b) vL is a maximum
(c) iL is a maximum.

(a) RL = Ω , vL = V , iL = A
(b) RL = Ω , vL = V , iL = A
(c) RL = Ω , vL = V , iL = A

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

Substituting for Il in eqn 2

Using equation 3 for both cases

case 1

case 2

solving equation 4 and 5 we get

[a]

when Vl * il is maximum then

Rl = Rs

⇒ Rl = 39.12 Ohms

Using equation 1

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

energy can cross the boundaries of a closed system in the form of heat and mass

explanation: