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 3 of 4.
Kunal said:
1 decade ago
Applying kcl in the circuit.
We will get,
I1 + 0.04 = 0.2.
We get I1 = 0.16.
We will get,
I1 + 0.04 = 0.2.
We get I1 = 0.16.
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.
Rstar said:
9 years ago
Apply superposition theorem both sources opposite direction.
0.2 - .04 = .16
0.2 - .04 = .16
L.Haque said:
1 decade ago
I1=.2A,when I1 active
I2=.04A,when I2 active
Final current=.2-.04=.16A(ans)
I2=.04A,when I2 active
Final current=.2-.04=.16A(ans)
Ahmad jaradat said:
1 decade ago
v=.04*68=2.72 (fixed)
so
v at new 68 = 2.72 (parallel)
i=2.72\68 = .04 a
so
v at new 68 = 2.72 (parallel)
i=2.72\68 = .04 a
Sravanthi said:
1 decade ago
Why it should be 0.1 A in each branch ? please explain it clearly.
Drishti said:
1 decade ago
By kirchoff's current law.
I +0.2=0.04.
I=0.04-0.2 = -0.16A.
I +0.2=0.04.
I=0.04-0.2 = -0.16A.
Vishal said:
1 decade ago
Jyoti can you explane how your gating 0.06 in last step?
Pramod thorat said:
1 decade ago
By Using current source theorems 0.2-0.04 = 0.16.
Shiva said:
1 decade ago
Apply kcl at centre node
0.2-0.04=1.6
0.2-0.04=1.6
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