The problem:

When the switch SW1 is in left position, ammeter AM1, with negliable inner resistance, shows current IA1=0.1A. When the switch SW is in right position, ammeter shows current IA2=0.15A. Find E2 by using superposition.

What I tried:

Try #1:
When I try using superposition and shut down the generator E1, I get U23=E2. When I shut down E2, I get U23=0. So by using superposition I get U23=E2+0=E2 but that doesn't help me find E2...

Try #2:
I tried compensating serial connection of E2 and ammeter with an ideal current generator Ic.
$$Ic=\begin{cases}IA1&SW1\leftarrow\\IA2&SW1\rightarrow\end{cases}$$ Then I tried finding the voltage between the nods 2 and 3 by using method of contour currents but all I get is U23=E2 or U23=0 as in previous case which is true but that doen't solve the problem.

How do I find E2 by superposition? When I try finding it, I constantly get E2=E2...


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ If I'm reading this correctly, when the switch is in the left position, both sources are active and the total current through AM1, due to both sources, is 0.1A. When the switch is in the right position, only E1 is active and the current is 0.15A. It seems to me that, by superposition, it must then be that the current due to just E2 is 0.1 - 0.15 = -0.05A. Finding the E2 that gives this current (with E1 zeroed) should be straightforward. \$\endgroup\$ Jan 17, 2018 at 3:41
  • \$\begingroup\$ @AlfredCentauri Yes you read it correctly. Just to make sure..does the current -0.05A go through the arc with E2 or with ammeter? \$\endgroup\$
    – Plexus
    Jan 17, 2018 at 12:48

2 Answers 2


When SW1 is in left position, Effect of both sources are there, I = 0.1 A. When SW2 is in right position, only the effect of E1 is there, I = 0.15 A. Also suppose SW1 is in the left postion and E2 is shorted, the resulting circuit is exactly the same as when SW2 in right position. It means the effect of E1 is again I = 0.15 A. Since the combined effect = 0.1 A for switch in left position, we can conclude that the effect of E2 is I = 0.1 - 0.15 = -0.05 A. If we assumed the direction of current to be +ve, if it flows from right to left of the ammeter, -0.05 A means the current is just in the opposite direction.i.e, from left to right of the ammeter. Now all left is to solve the resulting circuit using simple KVL or KCL.

enter image description here


If the switch is to the left (including E2 in the equations), then you have to short-circuit E1 once, and short-circuit E2 once, then add the current through the branch with E1.

When short-circuiting E1, you get the following schematic:


simulate this circuit – Schematic created using CircuitLab

I believe this is the annoying part, which is why I'm discussing it first. The matching KCL equations can be determined (current flowing in = current flowing out). The "trick" here is that we have a short-circuit by the amp-meter. We can resolve this by adding the unknown current \$i\$ (we define i as flowing out of v1 and into v2) and an extra equation to tie v1 and v2 together:




These give you three equations for three unknowns. You can calculate \$i_{only\ E_2}\$.

When shorting E2, you get the following schematic:


simulate this circuit

This is quite easy, as the total resistance is the series of two parallel resistors:

\$R_{tot}=(R_1//R_2) + (R_3//R_4)\$

\$i_{only\ E_1}=-\frac{E_1}{R_{tot}}\$

We can then apply superposition principle:

\$i=i_{only\ E_1} + i_{only\ E_2}\$

It turns out that when the switch is to the right, you get the same as our last circuit, so you can reuse that formula.

So when the switch is to the left: \$i = i_{only\ E_1}+i_{only\ E_2}\$

And when the switch is to the right: \$i = i_{only\ E_1}\$


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