I'm trying to answer following question:

In the ladder circuit in figure 3, find the current, I, if \$V_s = 7V\$.

Use the fact that the circuit is linear. This can be done by assuming that \$I=1A\$, and working out the potential drops back to the voltage source. Then use scaling.

I have easily computed the answer to be \$0.125A\$ using mesh analysis (and Maple). I cannot, however, figure out how to answer this question in the way suggested in the hint (exploiting the linearity of the circuit). Frankly it is something useful to know; solving 5 simultaneous equations during an exam would not go well.

Figure 3


If I understand the diagram, I is the current through the 1 Ohm resistor at the far right of the schematic.

What they're saying is, assume I is 1 A. Now you know, for example, the voltage at the next node left is 3 V. And you have a 1 A current through the 3-Ohm resistor. Since you have 2 A through the last two rungs of the ladder, you must have 4 V across the 2-Ohm resistor 2nd-from-the-right (see how much easier this would be if you just put designators in your schematic?). And so on, working back to the left.

Now when you get back to the source, you will get some voltage other than 7 V. Say you get 21 V (I haven't calculated, I just made up a number).

Since the system is linear, you know that if you scale this voltage down to 7 V, the current I will also scale down by factor of 3, so the actual I is 333 mA, not the 1 A you originally assumed.

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  • \$\begingroup\$ +1. This method works : a little mental arithmetic gives the same answer the OP quoted from Maple. If it helps ; "exploit the linearity" here means, at every step, treat the whole circuit to the right as a single resistor, whose value you calculated in the previous step. \$\endgroup\$ – Brian Drummond Feb 25 '13 at 17:26
  • \$\begingroup\$ @BrianDrummond, Your interpretation of "exploit the linearity" is how I started thinking about it. But I think what they are really talking about is the final step. Once you figure out what Vs would have to be to get 1 A at the last rung, you can scale linearly the voltage and current together to find out what the current is with 7 V at Vs. For OP, there's multiple ways to solve this problem (series/parallel combinations, mesh analysis, this "working-backward" method, etc.) All of them will come out with the same answer in the end. \$\endgroup\$ – The Photon Feb 25 '13 at 17:43

I like to redraw the circuit:

ladder circuit

Assuming currents all flow downwards through the resistors:

\begin{equation} i_{10} = i_{6} + i_{8} + i\\ i_{1} = i_{2} + i_{4} + i_{10} \end{equation}

Looking at the equivalent resistance of R9 and R11, we can tell that the same current flows through that branch as through R8 (2 + 1 = 3).

\begin{equation} i_{8} = i\\ i_{7} = 2 \cdot i\\ i_{10} = i_{6} + 2 \cdot i \end{equation}

Getting the equivalent resistance of R7, R8, R9, and R11. we can quickly compute this to be 3.5 ohms. Since this is half the resistance of R6, we know that twice as much current flows through R7 as through R6.

\begin{equation} i_{6} = \frac{i_7}{2} = i\\ i_5 = i_{10} = 3 \cdot i \end{equation}

Now let's get the equivalent resistance of R5, R6, R7, R8, R9, R10, and R11. We already know R7, R8, R9, and R11 is 3.5 ohms, and in parallel with an additional 7 ohms from R6 that gives us 7/3 ohms. Then add on the series R5 and R10 gives us 16/3 ohms.

Since the current will distribute between the two arms relative to their resistance,

\begin{equation} i_{4} \cdot R_4 = \frac{16}{3} \Omega \cdot i_5\\ i_{4} = \frac{4}{3} \cdot 3 \cdot i = 4 \cdot i\\ i_3 = i_4 + i_5 = 7 \cdot i \end{equation}

Repeating again, we get the equivalent resistance for resistors R3-R11.

\begin{equation} R_{3-11} = R_3 + \left(\frac{1}{R_4} + \frac{3}{16} \right)^{-1} = 2 + \frac{16}{7} = \frac{30}{7} \Omega \end{equation}

Now rinse and repeat the current distribution step.

\begin{equation} i_{2} \cdot R_2 = \frac{30}{7} \Omega \cdot i_3\\ i_{2} = \frac{6}{7} \cdot 7 \cdot i = 6 \cdot i\\ i_1 = i_2 + i_3 = 13 \cdot i \end{equation}

Now time to get the complete equivalent resistance:

\begin{equation} R_{eq} = R_1 + \left( \frac{1}{5} + \frac{7}{30} \right) ^{-1} = 2 + \frac{30}{13} = \frac{56}{13} \Omega \end{equation}

Then, using our basic Ohm's law:

\begin{equation} Vs = i_1 \cdot R_{eq} = 13 \cdot i \cdot \frac{56}{13} = 56 \cdot i \end{equation}

With Vs = 7V, \$i = \frac{7}{56} = \frac{1}{8} = 0.125A\$.

I didn't make the assumption that i = 1A, but if I did at the end I would have ended up with some Vs != 7V. Instead, I would have gotten Vs = 56V to get i = 1A.

Scaling can be done by dividing the target voltage by the scaled voltage, and likewise for the target current and scaled voltage:

\begin{equation} \frac{7V}{56V} = \frac{i_{actual}}{1A}\\ i_{actual} = 0.125A \end{equation}

Which is the same answer (as expected).

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  • \$\begingroup\$ And of course Photon already posted the answer, but I suppose I'm going to post this anyways because I spent some time working on this :P \$\endgroup\$ – helloworld922 Feb 25 '13 at 18:07
  • \$\begingroup\$ Thanks, I really like the rearranged diagram - very helpful. \$\endgroup\$ – Bart Platak Feb 26 '13 at 16:18

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