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lorihab1970
06.04.2020 •
Engineering
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
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Ответ:
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
Ответ:
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Explanation: