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

Question

^{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.

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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 ?

Answer 1

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.

Answer 2

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.