Electrical Engineering - Circuit Theorems and Conversions - Discussion
Discussion Forum : Circuit Theorems and Conversions - General Questions (Q.No. 8)
8.
A 12 V source has an internal resistance of 90 
. If a load resistance of 20 is connected to the voltage source, the load power, PL, is
. If a load resistance of 20 is connected to the voltage source, the load power, PL, isDiscussion:
5 comments Page 1 of 1.
Manoj said:
1 decade ago
A superconductor is a.
A. ) A material showing perfect conductivity and Meissner effect below a critical temperature.
B. ) A conductor having zero resistance.
C. ) A perfect conductor with highest di-magnetic susceptibility.
D. ) A perfect conductor which becomes resistance when the current density through it exceeds a critical value.
A. ) A material showing perfect conductivity and Meissner effect below a critical temperature.
B. ) A conductor having zero resistance.
C. ) A perfect conductor with highest di-magnetic susceptibility.
D. ) A perfect conductor which becomes resistance when the current density through it exceeds a critical value.
Yu Wei said:
3 years ago
By voltage divider, VL = 12(20) over 20+90 = 2.18V.
P = V^2/R,
PL = (VL)(VL)/RL. *others might mistake using Rth, we are only getting PL not PT so use RL,
PL = (2.18)(2.18)/20 = 237.62 mW => 238 mW.
P = V^2/R,
PL = (VL)(VL)/RL. *others might mistake using Rth, we are only getting PL not PT so use RL,
PL = (2.18)(2.18)/20 = 237.62 mW => 238 mW.
(4)
M.V.KRISHNA/PALVONCHA said:
1 decade ago
This is maximum power transfer theorem.
Rt=Rin+Rload
Rt=90+20=110ohm
It=Vs/Rt
It=12/110=0.109A
Pl=(It)^2*Rload
Pl=(0.109)^2*20
Pl=0.238W= 238mW
Rt=Rin+Rload
Rt=90+20=110ohm
It=Vs/Rt
It=12/110=0.109A
Pl=(It)^2*Rload
Pl=(0.109)^2*20
Pl=0.238W= 238mW
(2)
Don said:
6 years ago
Anyone, Please explain this answer.
(1)
Janmejay said:
1 decade ago
i=12/90+20
=a(say)
p=a^2*20
=a(say)
p=a^2*20
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