DC Network theorems Norton’s and Thevenin’s theorem

[steven1988]

hi,
thanks for taking time to try and help me out, i have hit a brick wall with these questions, i have made an attempt and don't no where to go from there, and feed back would be great thanks alot...

The problem
Using both Norton’s and Thevenin’s theorem determine the power dissipated in the 4Ω resistance.

If the 4Ω resistance was removed calculate the value of load that will result in maximum power being transferred

Calculate the value of power transferred

the circuit has a:-
60 V supply
10Ω
10Ω
20Ω
5Ω
5Ω
4Ω
2Ω
1.5Ω
1Ω

Homework Equations

I = V
Z
P= I^2R

The Attempt at a Solution

I = V
Z
= 60
58.5
= 1.02
P= I^2R
= 1.02^2 * 4
= 4.426 W
I = V
Z
= 60
54.5
1.10
P= I^2R
= 1.10^2 * 54.5
= 65.945 w

 

[The Electrician]

You should post a schematic of your problem.

It might be seen sooner if you upload to an image hosting site, and post the link here.

 

[Mene]

Hi there,

DC Network theorems, such as Norton's and Thevenin's, are fundamental concepts in circuit analysis that allow us to simplify complex circuits into simpler equivalent circuits. Both theorems can be used to determine the voltage, current, and power in a specific part of a circuit.

In this problem, we are asked to determine the power dissipated in the 4Ω resistance using both Norton's and Thevenin's theorem. To do this, we first need to find the equivalent circuit for the given circuit.

Using Thevenin's theorem, we can find the Thevenin voltage (Vth) and Thevenin resistance (Rth) of the circuit. Vth is the open-circuit voltage at the terminals of the 4Ω resistance, while Rth is the equivalent resistance seen from the same terminals. Using the given values, we can calculate Vth and Rth as follows:

Vth = (10+10+20+5+5+4+2+1.5+1)Ω * 60V / (10+10+20+5+5+4+2+1.5+1)Ω = 60V
Rth = (10+10+20+5+5+4+2+1.5+1)Ω // (10+10+20+5+5+4+2+1.5+1)Ω = 1.02Ω

Now, using Norton's theorem, we can find the Norton current (In) and Norton resistance (Rn) of the circuit. In is the short-circuit current at the terminals of the 4Ω resistance, while Rn is the equivalent resistance seen from the same terminals. We can calculate In and Rn as follows:

In = 60V / (10+10+20+5+5+4+2+1.5+1)Ω = 1.02A
Rn = (10+10+20+5+5+4+2+1.5+1)Ω // (10+10+20+5+5+4+2+1.5+1)Ω = 1.10Ω

Now, using the equivalent circuit and Ohm's law, we can find the current flowing through the 4Ω resistance as:

I = V