Electrical Engineering - Circuit Theorems and Conversions - Discussion

Discussion :: 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

[A]. 0 A
[B]. 1.2 A
[C]. 114 mA
[D]. 1.14 A

Answer: Option D

Explanation:

No answer description available for this question.

Sunil said: (Nov 24, 2011)  
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

Suraj 2000 said: (Aug 12, 2014)  
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.

Snowben Mascarenhas said: (Sep 30, 2014)  
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.

Jagadesh said: (Feb 23, 2016)  
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.

Srinuvasarao said: (Jul 19, 2016)  
I = Is * (Rs/Rs + Rl) = 1.2 * (12000/(12000 + 680)) = 1.14A.

Saroj said: (Nov 2, 2017)  
IL = 1.2 x (12000/12680).
=1.1356.
~1.14.

Calvince Omiti said: (Jun 23, 2021)  
RL = 12 X 10^3 ohms.
r=680 ohms.
I=1.2 A.
Solution
I load=1.2((680/(680+12000)),
I=1.14A.

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