# Is there a way to solve this circuit using superposition or KVL?

simulate this circuit – Schematic created using CircuitLab

Using KVL :
$-1 + 100(I_a+I_b) = 0$
$-1 + 100(I_a+I_b) = 0$
In both loops I get the same equation, so I cannot know how much current is drawn from each individual battery w/o using symmetry.

Using superposition:
Shorting either $Va$ or $Vb$ is a problem mathematically, so cannot proceed further.

I'm familiar with KVL, KCL and superposition so far, I can guess from symmetry $Ia=Ib$, but this is somehow not satisfying me. Is there a way to find the individual currents through the batteries mathematically w/o using symmetry ?

• related question Jun 10, 2017 at 4:46
• What's the current if Va doesn't equal Vb? Jun 10, 2017 at 16:09
• @sstobbe I think Vb would charge Va then; if Va was not a rechargeable type battery, it might explode. I can't think of a reason to connect two unequal voltage souces in parallel except for charging Jun 10, 2017 at 16:16
• Its a good question to think about but as drawn in violates what we assume for en.wikipedia.org/wiki/Nodal_analysis a node can't have two unique solutions for voltage Jun 10, 2017 at 16:19

## 2 Answers

There is no unique solution to this problem as presented.

In general, you should question whether the circuit model accurately reflects reality whenever you see two voltage sources connected in parallel. This example shows why you should worry about this even in cases where the two sources have the same voltage value.

If you add internal resistance to your source model, you will get a solvable problem.

Edit

In comments you said,

Assuming 0/same internal resistance for both batteries, I know each battery supplies half the current by symmetry. But I'm still wondering if its possible to arrive at this w/o using symmetry. I don't see any contradictions like 5 = -5 in my circuit ?

Your circuit doesn't have a stark contradiction like 5 = -5, but it is still unsolvable. The equation for your circuit is

$$I_a + I_b = 0.01 \rm A.$$

This equation has an infinite number of solutions, and mathematically there's no reason to think that the one with $I_a = I_b$ is in any way the preferred solution.

If you started with a circuit that includes internal resistance, and assumed the internal resistances of the two sources are exactly equal, and then took the limit as the internal resistance approaches 0, you would end up with that solution.

But it's really not a good idea to assume that internal resistance of two sources are exactly equal. This will most likely lead to making wrong conclusions about most realistic circuits.

• Ohk, so there is no way to figure out $I_a=I_b$ mathematically ? Using symmetry we can figure this w/o considering internal resistances right ? It seems we won't get solvable problem if the internal resistances of both batteries are same. In reality I might want to know how long each battery lasts. For this I need to know how much current is drawn from each battery. I know this is not a problem, for example, I see myself connecting two batteries in parallel to power a circuit so that the life of each battery is doubled(cont). Jun 10, 2017 at 5:13
• .. Assuming 0/same internal resistance for both batteries, I know each battery supplies half the current by symmetry. But I'm still wondering if its possible to arrive at this w/o using symmetry. I don't see any contradictions like 5 = -5 in my circuit ? Jun 10, 2017 at 5:16
• @Hiii, with the model as given, there's no reason to believe $I_a =I_b$. $I_a$ could be 1000 A, and $I_b$ would then be -999.99 A, with no contradiction. Your circuit simply has an infinite number of solutions, and nothing about it says that any of those solutions is any better than any other one. Jun 10, 2017 at 15:17

In this particular case, where the two sources are identical, with voltage, $\small V$, assume each has an internal resistance, r; do superposition or whatever; then let r tend to zero.

For the given circuit, This gives $\small I=\large \frac {V}{R}\small=\large \frac {1}{100}$, where $\small R=100\Omega$, and $\small I$ is the current through $\small R$.

In practice, the internal resistances of the sources will never be zero, so there will always be a solution, regardless of whether they're identical or not.