am having trouble with the following question. I have tried aimplifing the circuit, but get different answers when i do so. I instead started writing out equations for the queation. I am using that r_1 IS i_1. My equations are i_1-i_4-i_3=0 I_2-i_3-i_5=0 -10+v_1+v_3+v_5=0 V_4+v_5-v_3=0 V_6=v_5 V_6=v_7 -10+470i_1+4700i_3+2200i_5=0 10000i_4+10000i_5-4700i_3=0 4700i_6=10000i_5 4700i_6=2200i_7 Im just getting confused as what to do with all the equations now. So at i can obtain what the current and voltage is through each resistor
http://i.imgur.com/aHUFSix.png http://i.imgur.com/odCD6kC.png I try to solve your Circuits. I Get the Value At loop A 10V = 2.670k(i1) + 4.7k (i2) loop B 4.7k(i2) -10k(i1-i2) -10(i3) = 0V -10k(i1) + 14.7k(i2) -10(i3) = 0V Loop C 10k(i3) +4.7k(i4) = 0V Loop D 4.7k(i4) -2.2k(i1 -i2 - i3 - i4) = 0V -2.2k(i1) + 2.2k(i2) + 2.2k(i3) + 6.9k(i4) = 0 By solving in Determinants D = 4.29730834X10^15 D1 = 7.58902x10^12 (i1) = D1/D = 7.58902x10^12/ 4.29730834x10^15 = 1.7 mA There you go. Solve the other resistor ! ) Hope this is very helpful to you.
Using Mesh Current Method. The circuit: Equations: Mesh 1. 470(I1)+4700(I1-I2)+2200(I1)=10 Mesh 2. 4700(I2-I1)+10000(I2)+10000(I2-I3)=0 Mesh 3. 10000(I3-I2)+4700(I3-I4)=0 Mesh 4. 4700(I4-I3)+2200(I4)=0 Solution: You have a system of simultaneous equations. In order to solve such system you MUST have the number of equations equal to the number of unknowns. What are the unknowns? I1, I2, I3, I4. How many unknowns? 4. How many equations do we have? 4. Now go and solve it. If you don't know how to solve a system of simultaneous equations, Google it.
Intriguing, it was stated: what are the currents through all resistors? I find jumping straight to the current loops is the Mickey Mouse method. One should define all branch currents and voltages followed by using KCL and KVL to set up the relationships. It's paramount that the sum of all the depedent equations equate to zero before an analysis can begin. You will observe by making the equations independent (where their sum does not equate to zero),you're in a postion to choose any of the n currents in the n system of your choosing. By back substitution you can now solve for all branch currents and voltages. Master this method and it will be a sanity check for using other circuit theorems.
lol If you feel like really punishing yourself (and enjoy onset of headache), you could find the equivalent resistance of all the resistors, the circuit then becomes one voltage source and one resistor. Using Ohm's Law find the current. Then work backwards. Replacing equivalent resistance with it's components, one piece at a time. The benefit is that you find the branch currents and voltages as you work your way back to the original circuit from which you started. It is perfectly valid approach. Some exercises, in my experience, actually require you to do it that way. And while doing it for a circuit with 3 or 4 resistors is just time consuming. Doing it for a larger circuit is not, in my experience, enjoyable. I got a system of 4 simultaneous equations, which took two minutes to setup. A few more minutes to organize them. Another minute to put them into a calculator. BAM! I got 4 currents, 3 of them are the answers, and using them I find the other missing 3 currents. For the total of 6 currents. Now just use Ohm's Law to find voltages across individual resistors. Once that is done, you have all the currents, you have all the voltages, you are done.
Yes, you could use a variety of circuit theorems. So tell me, why could I neglect one of the internal loops and replace it with the external one instead?
Because you have 4 independent quantities and you have to use four independent equations to solve for them. There are at least ten loop equations you can choose from -- you just have to choice four of them that are independent. And you're not limited to loop equations, either. You have five node equations that you can throw in the mix. Then you have the equations for each component, which adds another eight or so. You have no shortage of equations. The trick is to choose a subset of those equations that are independent. That is where mesh current and node voltage analysis pays off -- they are techniques that automatically identify minimal sets of independent equations, incorporating the constitutive equations for the components as needed.