Electrical Engineering - Circuit Theorems and Conversions - Discussion
Discussion Forum : Circuit Theorems and Conversions - General Questions (Q.No. 4)
4.
A 680 
load resistor, RL, is connected across a constant current source of 1.2 A. The internal source resistance, RS, is 12 k. The load current, RL, is
load resistor, RL, is connected across a constant current source of 1.2 A. The internal source resistance, RS, is 12 k. The load current, RL, isDiscussion:
7 comments Page 1 of 1.
Calvince omiti said:
4 years ago
RL = 12 X 10^3 ohms.
r=680 ohms.
I=1.2 A.
Solution
I load=1.2((680/(680+12000)),
I=1.14A.
r=680 ohms.
I=1.2 A.
Solution
I load=1.2((680/(680+12000)),
I=1.14A.
(3)
Saroj said:
8 years ago
IL = 1.2 x (12000/12680).
=1.1356.
~1.14.
=1.1356.
~1.14.
(3)
Srinuvasarao said:
9 years ago
I = Is * (Rs/Rs + Rl) = 1.2 * (12000/(12000 + 680)) = 1.14A.
(2)
Jagadesh said:
10 years ago
As per current division rule to find current in any branch formula.
I = (Current*Resistance of opposite branch)/Total resistance of the two branches.
I = (1.2*12 k)/(12 k + 680).
I = 1.14 A.
I = (Current*Resistance of opposite branch)/Total resistance of the two branches.
I = (1.2*12 k)/(12 k + 680).
I = 1.14 A.
(1)
SNOWBEN MASCARENHAS said:
1 decade ago
Generally, A practical current source is represented by a current source in parallel with a resistor of high resistance, therefore in given problem the load resistor is parallel with the internal resistance. Therefore by applying current division rule we can obtain the exact result as 1.1356 Amperes which is approximately as 1.14 Amperes.
Suraj 2000 said:
1 decade ago
I think the answer should be 1.2 A as the source has an internal resistance of 12 K ohm and it is inherent and it will provide 1.2 A with this internal resistance. Hence the only resistance to be connected across should be 680 ohms and current value through series connection remains same.
Sunil said:
1 decade ago
Load resistor = 680 ohm,
Internal resistor = 12k ohm,
Source current = 1.2 A,
Current through load resistance
Irl = Is * (Rs/Rs+Rl)
= 1.2 * (12K / 12K+680)
=1.14A
Internal resistor = 12k ohm,
Source current = 1.2 A,
Current through load resistance
Irl = Is * (Rs/Rs+Rl)
= 1.2 * (12K / 12K+680)
=1.14A
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