Electrical Engineering - Circuit Theorems and Conversions - Discussion
Discussion Forum : Circuit Theorems and Conversions - General Questions (Q.No. 22)
22.
Referring to circuit given, if R1 is changed to a 68 
resistor, what will be the current through it?
resistor, what will be the current through it?
Discussion:
32 comments Page 1 of 4.
NIKHIL SINGH said:
6 years ago
Let current =i.
At any node sum of total current =0.
i+0.2-0.04=0.
Hence i= 1.6(leaving from node).
At any node sum of total current =0.
i+0.2-0.04=0.
Hence i= 1.6(leaving from node).
(1)
Vijay said:
1 decade ago
As both are current sources the current won't change at node hence.
Current in 120 ohm resistor = current in 68 ohm resistor.
I = 0.2-0.04 = 0.16A.
Current in 120 ohm resistor = current in 68 ohm resistor.
I = 0.2-0.04 = 0.16A.
(1)
Jagadesh said:
10 years ago
Simply the answer will be the difference of two current sources.
Why because the current from two sources enters the branch in opposite directions. So current through that branch will be I = 0.2-0.04 = 0.16 A.
What ever is the resistance placed in that branch the current through that branch will be 0.16 A.
Why because the current from two sources enters the branch in opposite directions. So current through that branch will be I = 0.2-0.04 = 0.16 A.
What ever is the resistance placed in that branch the current through that branch will be 0.16 A.
(1)
Ashish said:
9 years ago
Good explanation @L.Haque.
Shashi said:
1 decade ago
As per KCL,
Entering current = leaving current.
I1 = I2.
I1-I2 = 0.
0.2-0.04.
Entering current = leaving current.
I1 = I2.
I1-I2 = 0.
0.2-0.04.
Aestro said:
1 decade ago
I think 0.16 on -ve direction.
PANDI.R said:
1 decade ago
Simple ans:
R1 = 120, R2 = 68, IS1 = 0.2, IS2 = 0.04.
Change to R1 = 68.
So IS1 changed 0.04 then,
IS1 x IS2 = 0.2 x 0.2 = 0.16.
R1 = 120, R2 = 68, IS1 = 0.2, IS2 = 0.04.
Change to R1 = 68.
So IS1 changed 0.04 then,
IS1 x IS2 = 0.2 x 0.2 = 0.16.
Praveen said:
1 decade ago
If in the question current source are given then one thing I want to say that current source does not allow current through itself. So option B is correct.
KURAAKU NAGENDRA said:
10 years ago
Apply SPT,
For S1 active R1 = 0.2A(S2 = 0).
FOR S2 active r1 = -0.04A(s1 = 0).
Then I(total) = 0.2+(-0.04) = 0.16.
For S1 active R1 = 0.2A(S2 = 0).
FOR S2 active r1 = -0.04A(s1 = 0).
Then I(total) = 0.2+(-0.04) = 0.16.
Kotresh Bullannanavar said:
10 years ago
Only IS1 is active (And IS2 current travel to the ground).
Use current divider formula I2 = Im (R1/(R1+R2)).
= 0.2*(68/ (68+68)).
= 0.1 A.
Near is 0.16 A.
Use current divider formula I2 = Im (R1/(R1+R2)).
= 0.2*(68/ (68+68)).
= 0.1 A.
Near is 0.16 A.
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