oh, sorry my bad, -6A

OK. So what's the voltage at B, and then what is the current in the 6 ohm resistor?

 

Voltage at B = -6v
and I5 = 1A

This forum is not allowing me to make more than 5 posts in an hour. Can you explain this in one post please?

 

You've got to remember your previous results. The voltage at node C was calculated to be 12 volts. If the voltage at B is -6 volts, then what is the voltage across the 6 ohm resistor? Knowing that, calculate the current I5. You already know I7 and I6, so your new knowledge of I5 allows you to calculate I2. Knowing I2, you can calculate the voltage across the 2 ohm resistor which is connected to node C. You already know the voltage at node C, so knowing the voltage across the 2 ohm resistor allows you to calculate the voltage at node A. Knowing the voltage at node B and at node A allows you to calculate the voltage across Vs. Knowing the voltage at node A and node D allows the calculation of the voltage across the current source.

Do all this and post your results.

 

This is what I've got.

 

Your calculation for I5 is wrong. Since VB is -6 volts, I5 =( 12 - (-6) ) / 6 = 18/6 = 3A

Redo your calculations after this.

Not being careful with signs is probably one of the most common mistakes beginners make when solving networks.

 

VAD = -8v
Vs = 10v

 

VAD is not -8V, nor is Vs 10 V.

You're not showing some of your intermediate steps, so I don't know where you've gone wrong. Possibly you haven't noticed that the direction of I5 is not what is shown on the schematic. I5 goes to the left, so you may not have a correct value and direction for I2.

 

I'm going to lunch and won't be back for a few hours. Keep slogging away, and good luck.

 

 

Ok....What is the voltage at B? What is I4? What is I5?

 

Nazmul......Are you sure that you want to study electronics? These circuits are very, very simple.....you need to be able to do this kind of analysis in your head....as a second nature.

Later you will study reactance.......which is much more complicated. And takes of lot more math.

 

Nazmul, if you are using KCL you need to establish a convention for the sign of currents at a node. I like to use the convention that currents leaving a node are positive. The opposite convention is just as usable, but you have to choose one or the other and stick with it.

Assuming the currents leaving a node are taken to be positive, for node C what KCL tells us is that: I2+I5+I6+I7 = 0

Ignore the arrows showing current direction on the schematic and use the convention. Doing so, we have I7 = -1A because the current through the 8 volt source is entering node C; currents entering a node are considered negative by convention. I6 is 4A (positive) because it is leaving node C. I5 is 3A (positive) because it is leaving node C.

Given these values, we have I2+I5+I6+I7 = 0, or I2+3+4-1 = 0, or I2 = 1-3-4 = -6A. The minus sign means that I2 is 6A entering node C.

The arrows for I4, I5 and I7 on the schematic are pointing in the opposite direction of the actual current flow. Just ignore the arrows but keep the designations (names) for the currents.

If I2 is -6A, what is the voltage at node A, and what are the voltages Vs and VAD?

 

Nazmul......Are you sure that you want to study electronics? These circuits are very, very simple.....you need to be able to do this kind of analysis in your head....as a second nature.

Eventually, yes. But someone just learning this stuff can't be expected to be able to do them in their head right off the bat.

Having said that, by the time most (and notice that I say "most" and not necessarily "all") students start working with circuits of this kind, especially involving controlled sources, they are expected to have developed a reasonable level of proficiency with the basic circuit analysis techniques such as direct use of Ohm's Law, KVL, and KCL (which includes the whole issue of proper accounting for signs and voltage/current polarities) as well as their systematic application by means of Node Voltage and Mesh Current Analysis as well as Superposition and Thevenin and Norton equivalent networks.

So, Nazmul, it does seem like you are struggling with some concepts that you should have down pretty well by now. So you need to take a step back and make sure that you get this foundation firmly built, otherwise you will just be continuing to dig yourself deeper into a hole that will be increasingly difficult to get yourself out of.

I would suggest going back to the very beginning of your text and read it carefully and then work through EVERY exercise and problem until you are confident you have a strong handle on this. You will probably be surprised how quickly that can happen since you have been through it once already. Worst case, you might need to drop this course and either retake it or perhaps even the prerequisite course. It might be a tough pill to swallow, but it might also make the difference between completing your degree and not.