Electrical Engineering - Circuit Theorems and Conversions - Discussion
Discussion Forum : Circuit Theorems and Conversions - General Questions (Q.No. 10)
10.
Find the current through R1 in the given circuit.
Discussion:
9 comments Page 1 of 1.
Jigisha said:
7 years ago
Using Nodal analysis,
At node 1 -> v1/120 - 0.2 + (v1-v2/68) =0.
At node 2 -> (v2- v1)/68 + 0.04 = 0.
Thus solving simultaneously v1 and v2 equations we get v1= 19.2.
And so, at r2, I = 19.2/120 = 0.16.
At node 1 -> v1/120 - 0.2 + (v1-v2/68) =0.
At node 2 -> (v2- v1)/68 + 0.04 = 0.
Thus solving simultaneously v1 and v2 equations we get v1= 19.2.
And so, at r2, I = 19.2/120 = 0.16.
(1)
Sowjanya said:
8 years ago
Adding of two current sources but symbols are different
So. 0.2 -0.04 = 0.16
So. 0.2 -0.04 = 0.16
(1)
Rameshwar Choudhary said:
1 decade ago
Apply SUPERPOSITION THEOREM and solve it easily.
(i) Response due to Is1 is obtained by O.C. the Is2.
Which is equal to i1=+0.2A.
(ii) Response due to Source 2 is obtained by O.C, the source 1.
Which is equal to i2= -0.04A (* - sign to current direction in opposite to the source 1).
Complete response due to both source is obtained by SUPERPOSITION THEOREM.
i = i1+i2.
i = 0.2-0.04A.
i = 0.16A.
(i) Response due to Is1 is obtained by O.C. the Is2.
Which is equal to i1=+0.2A.
(ii) Response due to Source 2 is obtained by O.C, the source 1.
Which is equal to i2= -0.04A (* - sign to current direction in opposite to the source 1).
Complete response due to both source is obtained by SUPERPOSITION THEOREM.
i = i1+i2.
i = 0.2-0.04A.
i = 0.16A.
GIRIJANAND MESHRAM said:
1 decade ago
According to source transformation concept, the resistance R2 will
dummy resistance, because it is in series with the current source which is always determined as short ckt,.
Now both current source will be parallel to each other with opposite direction thus we get,
Current through R1 = (200*10^-3)-(40*10^-3).
= 160*10^-3.
dummy resistance, because it is in series with the current source which is always determined as short ckt,.
Now both current source will be parallel to each other with opposite direction thus we get,
Current through R1 = (200*10^-3)-(40*10^-3).
= 160*10^-3.
Praveen said:
1 decade ago
Just apply node analysis at node A.
Incoming currents = outgoing current.
I = I1+I2 now,
I1 = I-I2.
I1 = 0.2-0.04 = 0.16 amp.
Incoming currents = outgoing current.
I = I1+I2 now,
I1 = I-I2.
I1 = 0.2-0.04 = 0.16 amp.
(1)
SAURI said:
1 decade ago
Just apply..KCL.
we get 0.2 = 0.04+Ir1.
Ir1=0.16A.
we get 0.2 = 0.04+Ir1.
Ir1=0.16A.
M.V.KRISHNA/PALVONCHA said:
1 decade ago
From nodal analysis
(V/R1)-Is1+Is2=0
(V/R1)-0.2+0.04=0;
(v/R1)-0.16=0
required (V/R1)=0.16A;
(V/R1)-Is1+Is2=0
(V/R1)-0.2+0.04=0;
(v/R1)-0.16=0
required (V/R1)=0.16A;
Janmejay said:
1 decade ago
In other branch= I(s1)-I(s2)
Promod pandey said:
1 decade ago
We will apply kcl at node 1 we will get voltage than divide by 120.
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