Electrical Engineering - Circuit Theorems and Conversions - Discussion
Discussion Forum : Circuit Theorems and Conversions - General Questions (Q.No. 6)
6.
Determine IN for the circuit consisting of VS, R1, R2, and R3 shown in the given circuit.
Discussion:
20 comments Page 1 of 2.
Richard said:
1 decade ago
Guys from the nortons theorem that I know, In is supposed to be the current through the short circuit at AB.
That implies that is obtained from current through the 120ohm resistance which by current divider rule is.
(68/ (120+68) ) of the net current supllied by 75 voltage source.
Please correct me if I'm wrong!
That implies that is obtained from current through the 120ohm resistance which by current divider rule is.
(68/ (120+68) ) of the net current supllied by 75 voltage source.
Please correct me if I'm wrong!
Arundeep singh said:
1 decade ago
Rt = R1+ (R2*R3/R2+R3) = 68+ (68*120/68+120) =111.4 ohm.
It = Vs/Rt = 75/111.4 = 673 mA.
In = (R2/R2+R3) It = (68/68+120) 673 = 243.42 mA.
So answer B is absolutely right.
Rt=total resistance, It= total current, In = Norton's current).
It = Vs/Rt = 75/111.4 = 673 mA.
In = (R2/R2+R3) It = (68/68+120) 673 = 243.42 mA.
So answer B is absolutely right.
Rt=total resistance, It= total current, In = Norton's current).
Anand said:
1 decade ago
In is the Norton Equivalent current source, to find that :
Short the points A-B and then find the current drawn from the source. This gives the value of In. Answer comes out to be 673.22mA, close to A.
Short the points A-B and then find the current drawn from the source. This gives the value of In. Answer comes out to be 673.22mA, close to A.
(2)
ASHISH NAUTIYAL said:
1 decade ago
From nodal equation.
(v-75)/68+v/68+v/120 = 0.
By solving this we get v = 29.22.
Now by ohm's law to 120 ohm resistor.
IN = v/r = 29.22/120 = 0.244A.
or 1A = 1000mA.
SO, 0.244*1000 = 244mA.
(v-75)/68+v/68+v/120 = 0.
By solving this we get v = 29.22.
Now by ohm's law to 120 ohm resistor.
IN = v/r = 29.22/120 = 0.244A.
or 1A = 1000mA.
SO, 0.244*1000 = 244mA.
Seema said:
1 decade ago
Apply to KVL & solve it.
equations are:-
75-68I1-68I2=0.......(i)
-120I1+188I2=0.......(ii)
solve two equations....& same current flow in R3 & RL
i.e. I1-I2=242mA (Approx.)
equations are:-
75-68I1-68I2=0.......(i)
-120I1+188I2=0.......(ii)
solve two equations....& same current flow in R3 & RL
i.e. I1-I2=242mA (Approx.)
(1)
Karan saini said:
1 decade ago
Apply Thevenin Theorem-
finding Vth by nodal method
V-75/68+V/68=0
V=37.5
ten find Rth
Rth=(R1//R2)+R3
Rth=154ohm
In=37.5/154
In=244mA
Answer B is right
245mA is nearly 244mA
finding Vth by nodal method
V-75/68+V/68=0
V=37.5
ten find Rth
Rth=(R1//R2)+R3
Rth=154ohm
In=37.5/154
In=244mA
Answer B is right
245mA is nearly 244mA
Tes said:
1 decade ago
I think you should use thevnin theorem. That means find Vn by opening AB and find Rn by shorting AB. Finally divide Vn / Rn = In.
M.V.KRISHNA/PALVONCHA said:
1 decade ago
In=Vn/Rn.
Vn=(Vs*R2)/(R1+R2)
Vn=37.5v
Rn=R3+(R1//R2)
Rn=150 ohm
In=37.5/150
In=0.25A =250 mA (approx. equal to option B)
Vn=(Vs*R2)/(R1+R2)
Vn=37.5v
Rn=R3+(R1//R2)
Rn=150 ohm
In=37.5/150
In=0.25A =250 mA (approx. equal to option B)
Sankararao mamidi said:
1 decade ago
Apply nodal equation to the circuit.
[v-75]/68+v/68+v/120 = 0.
We get v = 29.22.
In = v/r.
In = 29.22/120.
In = 0.2435A.
[v-75]/68+v/68+v/120 = 0.
We get v = 29.22.
In = v/r.
In = 29.22/120.
In = 0.2435A.
(3)
Akshaykumar said:
6 years ago
Apply KCL to both loop, then solve those two equations, then the current i2 will be the answer, i.e=0.436A.
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