# Question 2_DC branch circuit analysis_AllaboutCircuits_worksheet

#### BC547

Joined May 22, 2018
40
Hello ,

I am trying to solve question 2 & 3 of the worksheet ( allaboutcircuits) . The link is here https://www.allaboutcircuits.com/worksheets/dc-branch-current-analysis/

However, I will post the question again. In the below circuit a dual source transistor circuit is modelled as a current controlled source with two voltages.
It is required to find the value of I using mesh current analysis.

When i solve for the left loop, my current is 6mA . For the right loop my current is -2.2mA which does not make sense. The right answer is 1.9mA.
I dont get it.

Besides, using this value the KVL in loop1 defies the law. (6V-1.9e-3*1e3) is not zero.

Any suggestions ?

#### WBahn

Joined Mar 31, 2012
26,398
You know that your answer is wrong because if I is 6 mA, how can (5 mA - I) be -2.2 mA?

You talk about "loop 1" but don't bother to tell us what "loop 1" is. Instead, we have to guess based on your equation but your equation doesn't walk around ANY loop in that circuit.

If you want us to help you find what you did wrong, you need to show us what you did in the first place.

#### BC547

Joined May 22, 2018
40
You know that your answer is wrong because if I is 6 mA, how can (5 mA - I) be -2.2 mA?

You talk about "loop 1" but don't bother to tell us what "loop 1" is. Instead, we have to guess based on your equation but your equation doesn't walk around ANY loop in that circuit.

If you want us to help you find what you did wrong, you need to show us what you did in the first place.

Thanks for writing!
The question is in the worksheet of this website (link is in original thread, Q2).

My approach,

For loop 1, i calculated the current based on KVL. Similarly for loop2. I know they should be same, but my analysis is incorrect which i am sure. The correct answer according to the writer is 1.9mA. I dont know how.
I have also attached my calculations.

Thx.

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#### WBahn

Joined Mar 31, 2012
26,398
The problem is that your loop equations do not account for the voltage drop across the 5 mA current source.

Think of the current source as being an adjustable battery whose output voltage has been adjusted to a value that results in it supply exactly 5 mA of current.

#### BC547

Joined May 22, 2018
40
The problem is that your loop equations do not account for the voltage drop across the 5 mA current source.

Think of the current source as being an adjustable battery whose output voltage has been adjusted to a value that results in it supply exactly 5 mA of current.
Thanks, now my current is -1.9mA. The direction will be reverse of that shown in figure. The voltage across the current source is 7.9V.
Is this the voltage drop across the emitter - collector in the transistor?

Thanks.

#### ericgibbs

Joined Jan 29, 2010
11,630
hi BC547,
I have checked the AAC maths, they are OK.

E
Hint: Note Vb

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#### WBahn

Joined Mar 31, 2012
26,398
Thanks, now my current is -1.9mA. The direction will be reverse of that shown in figure. The voltage across the current source is 7.9V.
Is this the voltage drop across the emitter - collector in the transistor?

Thanks.
Think about this. If the voltage across the current source is 7.9 V, is the current in the top left resistor flowing toward or away from the center junction? Is the current in the right resistor closing toward or away from the center junction? Are these consistent with the current flowing downward in the current source?

Again, show your work. We are not mind readers. We can't tell what you did wrong unless you show us what you did.

#### BC547

Joined May 22, 2018
40
Think about this. If the voltage across the current source is 7.9 V, is the current in the top left resistor flowing toward or away from the center junction? Is the current in the right resistor closing toward or away from the center junction? Are these consistent with the current flowing downward in the current source?

Again, show your work. We are not mind readers. We can't tell what you did wrong unless you show us what you did.
I committed mistake while accounting for the signs of the voltage drop across the current source. I got accurately current equal to 1.9mA and voltage drop of 4.1V. I have attached my calculation here.

Thanks for writing.

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#### WBahn

Joined Mar 31, 2012
26,398
Note that your indicated loop directions aren't consistent. In one your equation is summing voltage drops in the direction of the loop and in the other it is summing voltage gains. Either is find, but you should make the effort to be consistent. Otherwise you will end up making silly mistakes that could have been easily avoided.

You also should track your units throughout your work properly. As it is, you are just tacking on the units that you would like the answer to have. Instead, let the units speak for themselves. You are going to make mistakes -- we all do -- and most of those mistakes will mess up the units and scream out at you that a mistake has been made right after you make it. But they can only do that if you are properly using and tracking your units at each and every step. It's not that hard, takes very little additional time, and will quickly become second nature. It will save you LOTS of time and wasted effort in the long run.

#### BC547

Joined May 22, 2018
40
Note that your indicated loop directions aren't consistent. In one your equation is summing voltage drops in the direction of the loop and in the other it is summing voltage gains. Either is find, but you should make the effort to be consistent. Otherwise you will end up making silly mistakes that could have been easily avoided.

You also should track your units throughout your work properly. As it is, you are just tacking on the units that you would like the answer to have. Instead, let the units speak for themselves. You are going to make mistakes -- we all do -- and most of those mistakes will mess up the units and scream out at you that a mistake has been made right after you make it. But they can only do that if you are properly using and tracking your units at each and every step. It's not that hard, takes very little additional time, and will quickly become second nature. It will save you LOTS of time and wasted effort in the long run.
Hi,
I have followed the practice described in AAC textbooks on DC network analysis.
Besides, appreciate the feedback.
Thanks,

#### WBahn

Joined Mar 31, 2012
26,398
Hi,
I have followed the practice described in AAC textbooks on DC network analysis.
Besides, appreciate the feedback.
Thanks,
If you aren't interested in improving the quality of your work, getting higher grades, and spending less time doing work that you could have identified early on was guaranteed to produce a wrong answer, I can't stop you.