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.

676 mA
245 mA
431 mA
75 mA
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
20 comments Page 1 of 2.

Shekhar said:   1 decade ago
What is value of RL?

Kumar said:   1 decade ago
Tell me that how to calculate the In ?

Tirumala rao said:   1 decade ago
Solve the problem. I want explanation?

Hanumantha said:   1 decade ago
By nortan,s Theorem

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)

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!

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.)
(1)

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).

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

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.
(2)


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