Finding the currents and voltage using mesh analysis

Convert all the current sources into their equivalent voltage sources, replace capacitors with their DC equivalents and redraw the circuit and Find the currents I1, I2, I3, I4 and the voltage across R6 using the mesh analysis. (I was not able to draw the arrows for the current instead i have mentioned I1 to the right etc whereas it should be replaced with arrows.) Please hep me because i find this hard to do.

One way to label a current in a branch is using the name of one of its elements in series (if there are more than one element), for example, I(R1).
Then you may add "downwards", "upwards", "to the right" or "to the left" near the element to denote the assumed positive direction of the current.

For clarity, perhaps you may like redrawing the circuit then specifying the current to be found as, for example, I3 = I(R8) downwards.

I drew this circuit some time back and can you say me on how to draw it, using what method so that i can send you the image again. If not cant you reply me using this diagram itself, only fact is finding the currents I1, I2, I3, I4 and the voltage across R6 using the mesh analysis. I'm not clear on mesh but i have to definitely do this sum on today. Can u help me?instead of the words to the right. downwards you are to replace it with an arrow.

Sorry... It seems there are two I1 (one known and one to be calculated). Also there are two I2.

Do not consider the I1 and I2 marked next to the current source, it is just to denote the current sources.what you have to find is what i have mentioned as I1 to the right, I2 to he right, I3 and I4 and the voltage across R6

Do you mean we will work with these two current sources without giving them any name as it is done for all other elements in the circuits?!

All the elements in the circuit are named by some alphabet as R1,V1,C1 etc...but i made a mistake here by naming the current sources and the current passing by with the same alphabets, so you consider the current sources by some alphabet and proceed with the calculation by leaving the current flowing through the circuit as I1,I2,I3 and I4.Please help

As you know, the solution of any problem is done by steps.
The first step here is "replace capacitors with their DC equivalents and redraw the circuit". Could you do it?

I mean, do you know the DC equivalent of a capacitor?

Hint: if I4 is the DC current flowing in R7, I4=0.

I know the equation which is -j*i/2pifc is this equation correct. This question is itself a problem for me because i missed the particular class but i have to do this question deinitely. Please make sure to help me i really need the answers with the steps and i'll be able to configure the concept from your method and answers.

Sorry for the delay, my internet was out of service for a few hours.

Please note, the circuit here has DC sources only. It is a DC circuit that needs DC analysis.

In case of sinusoidal voltages and currents of frequency F, the impedance of a capacitor (C) is:
Vc/Ic = Xc = 1/jwC
where:
w = 2 * PI * F
1/j = - j (because j^2 = -1)
Here -j means that the voltage on the capacitor lags its current by a phase of 90 deg (PI/2 radian) which is equivalent to t=T/4 second where T=1/F.

Back to our DC circuit:
When a capacitor is fully charged (its voltage has reached its maximum, steady state), no more DC current passes thru it. In other words Idc_cap=0.
On post#8, I wrote (I4=0) because in its branch there is a capacitor (C2 in series with R7).
The same applies to the branch of C1 and R1. Their DC current is also zero.

So the first step is simplifying the circuit by removing the elements of every branch that has a capacitor. They are:
C2, R7, C1 and R1

I hope you got the idea of step one and you have time to redraw the smaller equivalent circuit.

I'm sorry for the late reply because just now i came online. What you have explained is clear and now i drew the circuit without the C2,R7,C1,R1 and simplified the circuit.i dont know how far i'm correct but please see to my answers and correct me. So this is my explanation - after removing the above elements from the circuit what was remaining were the 10mA, R2, R3, R5, R8, V1, V6, R6, R4, 12mA, and R9 and i redrew the circuit by only using these elements. And i further simplified the 12mA source and the R4 resistor which is parallel to it and replaced it with a voltage source of value 3.65*10^ -5. And i simplified the resistors R3 and R2 to a value of 545.45 ohm ( I considered R3 and R2 to be in parallel) .And now this 545.45 ohm resistor is in series with the 10 mA source and using this i converted it into a voltage source of value 5.45V, am i correct ? So in this case now the remaining elements are R6,R5,R8,R9,V2,V1 and the calculated voltage sources of 5.45V and 3.65*10^ -5V. So I4 will be Zero and now have to find only I1, I2, I3 and the voltage across R6 am i right? Please reply.

Since the mesh analysis should be used, the second step is to replace the current sources with their thevenin equivalents.

Could you find the thevenin equivalent of the source I1(10mA) and R1(2.2K)? Also of I2(12mA) and R4(0.33 Kohm)?

Please give the new voltage sources and resistors new names. This will help a lot in the following steps.

For instance, did you install a program to help you draw and update your electric/electronic circuits?

No,i dont have a software to draw circuits, if i'm to install one, how i'm i to do it. because then it would be easy for me to submit my answers to you and check them. And now to my question...First i'm to replace the capacitors with their DC equivalents that is by removing C1,C2,R1 and R7 and redraw the circuit without these elements, am i correct ? Now the next step is to consider current sources..for that you have said to find the thevenin equivalent that is you mean to find the VTh by considering each current source and their respective resistance? I cant understand on how to do this. Isn't what i have calculated above correct? That is for 10ma i've replaced with a voltage of 5.45V and for 12mA i have replaced with a voltage of 3.65*10^ -5v. I may be wrong but applying thevenin into this makes me confused, i mean i dont know to apply it here eventhough i sort of know about thevenin theorem because i have done some questions on that.Please help.

to find the thevenin equivalent that is you mean to find the VTh by considering each current source and their respective resistance?

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Yes, you are right.
Vth = I * R
where R is the parallel resistance of I.
and
Rth = R

As you know, the new Vth and Rth are is series.

Therefore:
Vth1 = V3 = I1 * R2
V3 = 10mA * 1K = 10 V
I named it V3 because there are already V1 and V2

Rth1 = R2 (as value)
In series with V3 we add the resistor Rth1 = R10 = 1K (the next name after R9, already in the circuit).

We repeat the same for I2 and R4.

Vth2 = V4 = I2 * R4
V4 = 12mA * 0.33K = 3.96 V

Rth2 = R4 (as value)
In series with V4 we add the resistor Rth2 = R11 = 330 Ohm.

Note: Don't forget to draw V3 and V4 with the proper polarity.

I use two free software; Kicad and LTspice. If necessary, I print the circuit schematic using the free limited version of "Bullzip PDF Printer" or by taking a screenshot.
Kicad is for drawing also the PCB layout from the schematic.
LTspice is to analyse the circuit (a circuit simulator).

Added:
I did for you the simplified circuit of step 2 (see attachement).
Is it possible to simplify even further this circuit?

Wow!You really did help me and thanks a million.To further simplify this circuit,can't we simplify the resistances R8 and R9 and consider it to be as one resistance value? This can be wrong because then we wont be able to find the proper I3 right? Now to find the currents I1,I2 and I3 am i consider individual loops? By the way how can i download the software you've stated?

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Wow!You really did help me and thanks a million.To further simplify this circuit,can't we simplify the resistances R8 and R9 and consider it to be as one resistance value? This can be wrong because then we wont be able to find the proper I3 right? Now to find the currents I1,I2 and I3 am i consider individual loops?

Step 3:
on step-2 circuit, we notice that the current (I1) of R3 is the same of R10 so we can combine them as R12 = ?
About the voltage on R6 (requested) could be deduced after knowing I2 since V(R6) = ?
And since R5 and R6 are also in series, they can be subistuted by R13 = ?
How many loops on the circuit we have now?

The reference node was selected and the three other nodes of the circuit were labeled V1, V2, and V3. Note that the node voltages are all with respect to the reference. So for example, V1 is the voltage with the "+" at the label V1 and the "-" at the reference node. The following equations can then be written:

Current leaving V1

(v1 - v4)/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0
Current leaving V2

(V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0
Current leaving V3

We don't know the current through the 12V nor through the 18V, and we cannot use Ohm's law to find it, since they are not resistors. So add unknown current labels to the circuit, as shown:

Now we can write the KCL equations:
iy + (V3 - V2)/5 - ix = 0
Current leaving V4

-iy + (V4 - V1)/2 = 0
Current leaving V5

ix + (V5 - 0)/1 = 0
Extra Equations

We now have five equations, but we have seven unknowns (the 5 unknown voltages plus 2 more unknown currents). We can get two extra equations -- just look at the voltage sources that caused the problems in the first place, where we had to add ix and iy. We'll write the voltage relationships between the node voltages:
V4 - V3 = 12V
V3 - V5 = 18V
Now we have 7 equations and 7 unknowns. To solve these equations, let's simplify it down to fewer equations. Solve for V4 and solve for V5 from those last two we added:
V4 = V3 + 12V
V5 = V3 - 18V
Substitute these two equations into the previous five, so that we eliminate V4 and V5. We now have 5 equations and 5 unknowns:
(v1 - [v3 + 12])/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0
(V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0
iy + (V3 - V2)/5 - ix = 0
-iy + ([V3 + 12] - V1)/2 = 0
ix + ([V3 - 18] - 0)/1 = 0
Continue using substitution (Gaussian elimination) to narrow it down. Next, solve for ix in the last equation:
ix = 18 - V3
and substitute back to get 4 equations and 4 unknowns:
(v1 - [v3 + 12])/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0
(V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0
iy + (V3 - V2)/5 - [18 - V3] = 0
-iy + ([v3 + 12] - V1)/2 = 0
Next, solve for iy in the last equation:
iy = ([v3 + 12] - V1)/2
and substitute back to get 3 equations and 3 unknowns:
(V1 - [V3 + 12])/2 + (V1 - V2)/3 + (V1 - 0)/2 = 0
(V2 - V3)/5 + (V2 - 0)/4 + (V2 - V1)/3 = 0
([V3 + 12] - V1)/2 + (V3 - V2)/5 - [18 - V3] = 0
Now multiply each equation by a constant to clear out fractions. That is multiply by the Least Common Denominator (multiply first by 6, second by 60, third by 10).
3 * (V1 - [V3 + 12]) + 2 * (V1 - V2) + 3 * (V1 - 0) = 0
12 * (V2 - V3) + 15 * (V2 - 0) + 20 * (V2 - V1) = 0
5 * ([V3 + 12] - V1) + 2 * (V3 - V2) - 10 * [18 - V3] = 0
Next, multiply through
3*V1 - 3*V3 - 36 + 2*V1 - 2*V2 + 3*V1 = 0
12*V2 - 12*V3 + 15*V2 + 20*V2 - 20*V1 = 0
5*V3 + 60 - 5*V1 + 2*V3 - 2*V2 -180 + 10*V3 = 0
Now gather like terms:
8*V1 -2*V2 -3*V3 = 36
-20*V1 +37*V2 = 12*V3 = 0
-5*V1 -2*V2 + 15*V3 = 120

now it is complete.



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Sorry Joyson1988, on which circuit are your caculations applied?
For example, I couldn't find the 18V.

We don't know the current through the 12V nor through the 18V

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kerim, now i'v got three loops, am i correct?

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kerim, we cant now itself deduce R6 and R5 before finding V across R6 right?

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Can you say whether my equations are correct? 10=I1*2.2k ohm + I3*2k ohm and 12-3.96=I2*(R5+R6+1/R11)

You are right. There are 3 loops.

Let us write the equation of every loop by using symbols only (not values and as shown on the simplified step-3 circuit, attached below).

Note: The loop currents are I_1, I_2 and I_3 while I1, I2 and I3 are as specified in the original problem.