# Batteries in parallel circuit

• Ezio3.1415

#### Ezio3.1415

I want to know how we can find out the net voltage of a circuit where two batteries with different voltages are connected in parallel... Let's consider the problem stated below...

If I have a circuit where there are 2 batteries in parallel(5V & 15V)... And it is connected in series with another 10V battery... Each battery has an internal resistance of 1 ohm... (The batteries are rechargeable so current can flow either way through them)

What is the net emf in this circuit?

Think of a 5 V battery as a pair of nodes, one at 0 V and the other at 5 V.

What you are doing when you connect a 15 V and 5 V battery in parallel is specifying that the two ground nodes (0V and 0V) are connected, which is fine, but also that the two other nodes, +5V and +15V, are connected. By a conductor. This doesn't make sense because the connectors in the circuit should be at the same potential.

In practice, you'd get a reverse current flowing through your weaker battery, which is not recommended!

Moral of the story is to only ever connect the same types of batteries in parallel.

A current of 5 amps will flow through the mismatched batteries, so there will be an EMF of 20 volts for the circuit. The 15 volt battery will rapidly discharge. A 5 amp current is so large, it could cause a fire.

So do the two batteries in parallel averages each other out? (15+5)/2=10

Yes, but you'll ruin both batteries and may start a fire.

So do the two batteries in parallel averages each other out? (15+5)/2=10

Not in general, no. Suppose, for instance, that you have a 12 volt automotive battery wired in parallel with a string of 4 AAA cells. That's 12 volts and 6 volts. You would not expect the voltage to be 9 volts. You would expect the automotive battery to overwhelm the AAA cells.

One reason for this is that the internal resistance in the string of AAA cells is likely to be much higher than that in the automotive battery. Ball park resistance in a 4 cell chain of AAA's is about 0.4 ohms. Ball park resistance in an automotive battery would be around 0.002 ohms.

Six volts (net) divided by 0.402 ohms (total) would give a current of around 15 amps.

So the current across the AAA cells is 6 volts (their nominal rating) plus 15*0.4 = another six volts associated with the current flowing backward across the internal resistance. That's a net of 12 volts minus a tad across the AAA array.

And the current across the automotive battery is 12 bolts (its nominal rating) minus 15*0.002 = .03 volts associated with the current flowing forward across the internal resistance. That's a net of 12 volts minus a tad across the automotive battery.

Can you please find the current through 6V,10V & the 12 V battery here?
That would really clarify things for me...

And in case of solving the problem, I will use kirchhoff law...

What I am asking is what happens exactly at first and what follows later?

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Can you please find the current through 6V,10V & the 12 V battery here?
That would really clarify things for me...

And in case of solving the problem, I will use kirchhoff law...

What I am asking is what happens exactly at first and what follows later?

So you have 6 volts at 0.4 ohms, 12 volts at 0.002 ohms and 10 volts at 5 ohms all wired in parallel. It might (or might not) help to draw the picture with each battery in series with its corresponding internal resistance.

One way of solving the new system is to set up a series of simultaneous equations. Let us arbitrarily set the potential of the bottom connector to zero and call the potential of the top connector v. Denote the current flow from top connector to bottom across the three batteries by c12, c10 and c6 respectively. Assume that the positive pole of each battery is toward the top so that v will be positive. Adopt a sign convention so that current flowing from top to bottom is positive.

Then we have:

c12 + c10 + c6 = 0
c12 = (v-12) / .002
c10 = (v-10) / 5
c6 = (v-6) / .4

That's four linear equations in four unknowns. That's solvable.

Another way to proceed would be to convert the first pair of batteries to a Thevenin equivalent and then add that to to the remaining battery. Let's try that.

Start with the 6 volt and the 10 volt battery in isolation...

This part of the circuit can be viewed as a pair of resistors and a pair of batteries wired in series in a loop The total resistance in the loop is 5 ohms plus 0.4 ohms = 5.4 ohms. The total voltage in the loop is 10 - 6 = 4 volts since the batteries oppose one another. The current flowing in the loop is 4 volts / 5.4 ohms ~= .74 amps.

Consider the current flowing across the 6 volt battery. That's 0.74 amps across 0.4 ohms = .296 volts. That's in addition to the 6 volts from the battery for a total of 6.3 volts.

Consider the current flowing across the 10 volt battery. That's 0.74 amps across 5 ohms = 3.7 volts. That's subtracted from the 10 volts from the battery for a total of 6.3 volts. [The fact that this matches is a nice double-check on our work].

So the 6 volt and the 10 volt battery has a Thevenin Equivalent comparable to a 6.3 volt battery wired in series with some resistance. The resistance will be equal to that of a 5 ohm and a 0.4 ohm resistor wired in parallel. That's 1 / ( 1/5 + 1/0.5 ) = 1 / 2.7 ~= 0.37 ohms.

Now we have a simplified circuit with a 12 volt battery and a .002 ohm resistance wired in series with our Thevenin equivalent of a 6.3 volt battery with a 0.37 ohm resistance. Again, the batteries oppose each other. That's a total resistance of 0.372 ohms and a total voltage of 5.7 volts through the loop. The current flowing will be 5.7 volts / 0.372 ohms = 15.3 amps.

15.3 amps across across .37 ohms is 5.67 volts added to the 6.3 volts of the Thevenin equivalent for a total of 11.97 volts.

15.3 amps across .002 ohms is .0306 volts subtracted from the 12 volts of the automotive battery for a total of 11.97 volts. [Again, this is a nice confirmation]

Now, the question that was asked was the current across the three batteries.

12 volts minus 11.97 volts = 0.03 volts. Divide by .002 ohms to get 15 amps forward across the 12 volt battery.

10 volts minus 11.97 volts = -1.97 volts. Divide by 5 ohms to get 0.394 amps backward across the 10 volt battery.

6 volts minus 11.97 volts = -5.97 volts. Divide by 0.4 ohms to get 14.925 amps backward across the 6 volt battery.

I'm willing to chalk up the fact that these don't exactly add to zero to the round-offs that I did through the calculation. [The 0.03 volt draw-down on the 12 volt battery, in particular, is sufficiently imprecise to lead to the discrepancy].

Now that's what happens initially, assuming that all the batteries are acting ideally and all of the internal resistances are nice and linear. As things proceed, the 6 volt battery is being over-charged at a hideous rate. Those cells will likely explode. If driven at a sufficiently low rate to avoid that fate, the result might instead to be to increase the voltage provided by the 6 volt battery due to changes in the concentrations of the reactants inside the cells. But that gets you into the realm of electrochemistry and has nothing to do with Kirchoff's laws.