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 2 of 2.
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)
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
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).
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)
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!
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)
Hanumantha said:
1 decade ago
By nortan,s Theorem
Tirumala rao said:
1 decade ago
Solve the problem. I want explanation?
Kumar said:
1 decade ago
Tell me that how to calculate the In ?
Shekhar said:
1 decade ago
What is value of RL?
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