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This is a pretty simple question really and I'll feel embarrassed when I get the solution but its bugging me for now.

enter image description here

The goal is to solve for I going from left to right. No values given for any of the resistors. My instinct was to do mesh current and I came up with this solution.

enter image description here

I'm not using I2 at all and that's bugging me intuitively. The bottom part of the circuit is essentially being ignored and that doesn't seem right. Should I be doing I1 - I2? Is there a better approach to solving this?

Thanks

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  • \$\begingroup\$ Not too sure about the question, are you asking how to calculate the current through R3, R4 in parallel? \$\endgroup\$ – lucas92 Jan 30 '17 at 19:47
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    \$\begingroup\$ Draw a third loop which contains the voltage source with current I3. \$\endgroup\$ – Tut Jan 30 '17 at 19:57
  • \$\begingroup\$ Replace the short circuit with a resistor, Rx; do the loop analysis; then let Rx = 0 \$\endgroup\$ – Chu Jan 30 '17 at 23:29
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Like the comment above suggested, if using mesh analysis, you need a third loop that has a voltage source from Vdd to ground. Then you can solve for the three loop currents and subtract I2 from I1 to obtain the solution. The algebra is a bit messy. A perhaps simpler approach is to convert the left and right sides to their Thevenin equivalents which reduces the problem to a single loop:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ This makes sense. Couple questions. Does this assume that there's two different voltage sources or are you just representing the same one? Do I then just do mesh analysis on this one loop for the answer? \$\endgroup\$ – samz_manu Jan 31 '17 at 1:07
  • \$\begingroup\$ Maybe it's already too simple to see: For a single mesh circuit you don't have to do any sophisticated analysis: combine the (two) sources to one, combine the (two) restors to one. Now you have a circuit of one source connected to one resistor. Now it shouldn't be a problem to find the current. \$\endgroup\$ – Curd Jan 31 '17 at 11:07
  • \$\begingroup\$ Referring to the original circuit, R1 and R3 are both at the same voltage. From an analysis viewpoint, it doesn't make any difference if you have a single independent voltage source connected to both R1 and R3, or one source connected to R1 and a second source connected to R3. Using separate sources allows you to convert the left and right sides to their Thevenin equivalents. To solve for the current, just divide the sum of the voltages, Vdd R2(R1+R2) - Vdd R4(R3+R4) by the total resistance, R1 in parallel with R2 plus R3 in parallel with R4 \$\endgroup\$ – user28910 Jan 31 '17 at 14:56
  • \$\begingroup\$ Oops, that should be Vdd times R2 divided by (R1+R2), etc. \$\endgroup\$ – user28910 Jan 31 '17 at 15:17
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A voltage source plus a serial resistance can be converted to a current source plus a parallel resistance. This will greatly simplify your analysis.

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  • \$\begingroup\$ "... cannot be converted to a current source" - don't you mean "... CAN be converted...". \$\endgroup\$ – Kevin White Jan 30 '17 at 23:45
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I did it this way:

schematic

simulate this circuit – Schematic created using CircuitLab enter image description here

Where:

enter image description here

This is rather messy but does solve it.

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  • \$\begingroup\$ That doesn't solve it. What are I1 and I2 in terms of VDD and the resistors? \$\endgroup\$ – Chu Jan 31 '17 at 0:31
  • \$\begingroup\$ I've added terms for I1 and I2 now, thanks \$\endgroup\$ – wevcore Jan 31 '17 at 9:43

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