I'm having trouble finding the voltage between the points 0 and 1. Can anyone help me? By applying KVL on each closed loop I find that Vi+IiRi is constant for every "i" but I don't know how to find the current flowing trough each resistor and therefore the corresponding voltage.

  • \$\begingroup\$ Are the 1 and 0 voltages or just designators of nodes? \$\endgroup\$ – horta May 29 '14 at 17:08
  • \$\begingroup\$ they are only names given the the nodes \$\endgroup\$ – user3479115 May 29 '14 at 17:10
  • \$\begingroup\$ The points are actually two ends of a parallel circuit. As with any ends of a parallel circuit, voltage is a constant. \$\endgroup\$ – arjun May 29 '14 at 17:36
  • \$\begingroup\$ en.m.wikipedia.org/wiki/Millman's_theorem \$\endgroup\$ – Bitrex May 30 '14 at 6:44
  • \$\begingroup\$ The sum of all (Vn-V(1) )/Rn = 0 \$\endgroup\$ – Russell McMahon May 30 '14 at 8:54

After seeing Horta's answer, I got another way to do it.

Replacing each branch with its Norton's equivalent, we get




Where, $$I_x = \sum_{i=1}^{n}{\frac{V_i}{R_i}} \mathrm{\ \ and\ \ }R_x = R_1||R_2||\cdots ||R_n$$

And hence voltage between \$1\$ and \$0\$, $$V_{1,0} = R_x\times I_x = R_x \times \sum_{i=1}^{n}{\frac{V_i}{R_i}}$$


Using node analysis: $$ \frac{V_1 - V_p}{R_1} + \frac{V_2 - V_p}{R_2} ... = 0 $$

$$ \frac{V_1}{R_1} - \frac{V_p}{R_1} + \frac{V_2}{R_2} - \frac{V_p}{R_2} ... = 0 $$

$$ V_p(1/R_1+1/R_2...) =\frac{V_1}{R_1}+\frac{V_2}{R_2} ... $$

$$ V_p = (\frac{V_1}{R_1}+\frac{V_2}{R_2}...)\times\frac{1}{(\frac{1}{R1}+\frac{1}{R2}+...)} $$

$$ V_p= R_x\times\sum\limits_{i=1}^n \frac{V_i}{R_i} $$ where \$V_p\$ is the voltage at point 1 and \$R_x = R_1||R_2||\cdots||R_n\$

Further simplification from here eludes me atm.

With the sum notation it also starts looking like an average which makes sense. You're ending up with an average voltage weighted by the resistors for each section.

  • 2
    \$\begingroup\$ In the fourth equation, shouldn't it be 1/(1/R1 + 1/R2 + ...) instead of (R1 + R2 + ...)? \$\endgroup\$ – user36113 May 29 '14 at 17:50
  • \$\begingroup\$ Ah, good catch, you're right. Edited it to reflect. \$\endgroup\$ – horta May 29 '14 at 18:00

My brother had a garage full of lights with varying currents due to varying resistance. But instead of being all on, they were all similarly dim. That was because: he had them all series connected! A series circuit is R1 + R2 +R3 ... The lights should have been in parallel, so each low lamp resistance would swamp out the almost 0 source resistance. Lesson: parallel circuits with a common source share the current, which in total is the same at the breaker end flowing out, as it is, in total, flowing into ground -- series circuits share the voltage; parallel ones share the current.

But what if he had connected them as in this problem? If all the voltages were different, currents would depend on the values of the resistors. The total of the currents would be V1/R1 + V2/R2 ... Since the resistors are connected only to each other at one end, what is the current doing?

Each voltage source higher in value than the one connected to the lowest resistor Rmin, will dump enough current down its resistor, so that its resistor drops the difference between its voltage source and the source at Rmin. If a voltage source is lower in value than Rmin's source, the current will go the other way, and the resistor connected to the lower voltage will again pass enough current, to account for the voltage difference. Since Rmin will always get enough +ve current from its voltage source (and just enough) to drop its own voltage source's voltage V@Rmin, down to 0v, not much current (=0A?) from anywhere else, other than Rmin's voltage source, passes through Rmin.

This will be controlled by all the voltage sources acting together to drop their voltage, minus Rmin's voltage, across their resistor, no matter in which direction current goes through their resistor, to or from point 1. If a voltage source is higher than the one connected to Rmin, the current will be towards Rmin. If the V connected to a non-Rmin resistor is lower than the Rmin's, current flows into it. Since all Vs are voltage sources with respect to their negative terminals at point 0, a virtual ground, the voltage across Rmin will always be the voltage of its own source, no more or less. The voltage is controlled by that resistor. The other resistors will each source or sink just enough current to account for the difference between the two voltage sources WRT point 1, when Rmin is passing exactly the positive current provided by the positive voltage connected to it.

All the voltage sources combined assure that that is the case, providing or sinking current depending on whether they are higher or lower than the voltage connected to Rmin. All the currents from or to all resistors besides Rmin, add up algebraically, to 0. Their polarities depend on their source voltage with respect to the common voltage at point 1.

In series circuits, the common current through all resistors, added together, causes the single voltage across all of them to be divided according to their relative values.

In circuits like this one, different voltages across each resistor, and their common connections, the total current divides according to the relative R values (polarities determined by voltages at each end relative to the other). Of course if voltages are equal no current flows.

In parallel circuits no current can be higher than Rmin's. Lower voltages than Rmin's pass current away from point 1. Higher voltages than Rmin's pass current towards point 1. Amount depends only on the non-minimum resistor value, and the voltage difference, regardless of direction. Polarity depends on voltage relative to point 1.

Voltage at point 1 == V@Rmin.


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