Electronic – the fraction of current taken by this resistor

circuit analysisthevenin

A network of resistors has a pair of input terminals AB connected to a DC supply and a pair of output terminals CD connected to a load resistor of 120 Ohm. The resistances of the network are AC=DB=180 Ohm and AD=BC=80 Ohm.

What is the ratio of the current in the load resistor to that taken from the supply?

NB: Labels marked in red are for the mesh analysis I've made!

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The question is probably not very hard, however I am trying to find the answer by not applying Norton's or Thevenin's theorems, but only Kirchhof's laws and therefore Nodal and Mesh analysis.

What I've tried so far:

delta-star transformation With that it's easier to find the total resistance, but that's it.

schematic

^{simulate this circuit}

nodal analysis ended up with (with reference to the initial diagram)

$$\frac{C-A}{180} – \frac{B-C}{80} = \frac{D-C}{120}$$
$$\frac{D-C}{120} = \frac{B-D}{180} – \frac{D-A}{80}$$

It seems be a deadlock with to many unknown values.

Finally mesh analysis, after solving simultaneous equations assuming current to flow clockwise in every closed loop (DC source connected in between points A and B) I've got following.
$$I_{1} = \frac{V}{120}, I_{2} = \frac{V}{200}, I_{3} = \frac{V}{300}$$

UPD: Numerical answer for this question is (negative direction) 5 and 0.2 (for positive direction of current flowing), but the most important is how this answer was derived and ultimately this is my question.

Thanks in advance!

Best Answer

The current through the load resistor can be obtained by finding the THEVENIN EQUIVALENT across it.If you are unaware about do check out thevenin intro or any other online material on it.It is very useful in solving networks for currents and voltages.

The thevenin equivalent across 120 ohm is :

From the above circuit the current through load can be easily calculated as:

I(Load) = [ 0.385Vs /(110.76 + 120) ] = 0.0016684 Vs ( where Vs is the supply voltage at input)

Now, to get the current drawn from the supply voltage Vs at the input,

Step 1: Find the equivalent resistance across the input source which involves star to delta conversion as you have mentioned.

The R(equivalent) = [ 37.9 + ( (56.8 + 80) || (25.3 + 180) ) ] = 120 ohms.

Step 2:

The current drawn from the input is : I(supply) = Vs/ Req = 0.00833 Vs.

Now that we have both currents in terms of the input supply voltage,

The ratio of load current to the current drawn from the supply is:

I(Load) / I(supply) = 0.001664 /0.00833 = 0.2

If the answer is 5 then it should be the inverse of the above as current through load is always less than supply current.

Edit: Since you wanted to solve it by mesh analysis here is an easier way:

Related Solutions

Electronic – How to decide which method to use in circuit analysis

Basically there are few different things here that appear lumped together in your mind and which may be better seen separately.

First you have the classical methods for solving circuits, such as the mesh analyses, node-voltage analyses, full equation system, reduced equation system and maybe some more I forgot.

First, the full equation system has no excuse to be used, so don't use it. The reduced equation system should only be used for very simple circuits where the preparation to use one of the two other systems would take more work than solving the circuit using reduced system. After some practice, you should be able to determine how much work you'll need to solve the circuit using each system (so try using all 3 systems on some problems, just so you can compare how much work each requires).

Next, you should have good theoretical knowledge of how each system works. The circuit may be such that some system cannot be easily used with it. You should also note the special cases where the mesh analyses and nodal analyses are especially useful. For example, in mesh analyses, if you have lots of constant current sources, you get lots of trivial equations for mesh currents. Same thing with nodal analyses and voltage sources. So basically, you should calculate how many equations you'll get with each and pick one with least equations, but do take into account special cases of current source and voltage sources when calculating the number of equations!

After that, you have the theorems which you use. I'd say that you should first see if any theorems can be applied to a circuit and only after applying them try to solve it using the systems. For example, you should use Norton/Thevenin when you have a big circuit where one for few elements change. For example, you have a circuit with a rheostat and you need to calculate say current through it on various settings. In this case, just replace the rest of the circuit with Thevenin's generator, since it doesn't change. In case of superposition theorem, it's useful in cases where you have sources that turn on and off and you should calculate their effects on some part of the circuit. Same thing goes for other theorems, like bisection (where you have symmetrical circuit, so you only need to solve one half of it) or compensation (which is useful when you have sources whose values change).

So for theorems, the general idea is to find use cases where each of them will actually allow the circuit to be simplified. So when you have a problem, ask yourself not "How am I going to solve this using method X?" but "Why am I going to solve this using method X?". This should work even on problems in textbooks where they are divided by areas. So as I said before, try solving one problem using several different methods. See which ones can be applied to the problem, which ones can't be applied to the problem, ask yourself why for each and then take a look and see which method is the most optimal (in sense that the least number of equations needs to be solved or that you get a significant number of simple equations) and try to understand why the most optimal method is the most optimal method. This way, you'll see when it's counterproductive to apply some theorem, when you gain nothing by applying some theorem and when applying a theorem actually helps. Same story goes for reduced system, nodal and mesh analyses too.

Need help finding the mistake using Thenevin’s Theorem for this circuit (not homework!)

Calculation mistake at Step number 4. You wrote:

I2 = 24/15K = 8/3K A = 2.667 mA

But 24/15k = 8/5k not 8/3k. So I2 is

I2 = 24/15k = 8/5k A = 1.6mA

You will get Vth = 25.6 V.

Best Answer

Related Solutions

Electronic – How to decide which method to use in circuit analysis

Need help finding the mistake using Thenevin’s Theorem for this circuit (not homework!)

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