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 2 of 4.
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)
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.
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.
Ayman said:
1 decade ago
Why the current of the 0.2 amperes does not divide into two equal currents at the point between 68Ohm resistors, should not it?
RAJIV SINGH said:
9 years ago
By current division rule = 120 * 0.2/(120 + 68) = 0.12. So total current through 68ohm resistor is 0.12 + 0.04 = 0.16.
KaushalSingh said:
1 decade ago
Condition given that R1=68 and R2=68.
Apply KCL at central node.
So current flows equally to both the resistors.
Apply KCL at central node.
So current flows equally to both the resistors.
Willstone said:
9 years ago
Current does not change in the 120 ohm line. If resistance change, then only voltage drop across it changes
Ross said:
9 years ago
Here resistance value doesn't affect current magnitude as it will be the difference of 0.2 and 0.04 = 0.16.
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)
Pradeep said:
1 decade ago
Apply kcl v1/68=2-.04.
v1/68 = 0.16.
v1 = 0.16*68.
v1 = i1*r1.
0.16*68 = i1*68.
i1 = 0.16A.
v1/68 = 0.16.
v1 = 0.16*68.
v1 = i1*r1.
0.16*68 = i1*68.
i1 = 0.16A.
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