Electronic – Analysis linear ladder circuit


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

Best Answer

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.