# current produced by two batteries connected in parallel

Discussion in 'General Electronics Chat' started by PG1995, Oct 31, 2011.

1. ### PG1995 Thread Starter Active Member

Apr 15, 2011
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Hi

Please have a look on the attachment. You can find my questions there. Please help me with them. Thank you.

Regards
PG

PS: The batteries in Q2 have the same emf, say, 5V.

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2. ### ErnieM AAC Fanatic!

Apr 24, 2011
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Q1: Correct, the net current is due to the 10V battery. But current may be defined in any random mannor before you do any analysis. Here, if the current is defined as running left to right at the top of the circuit, you just have a current of negative magnitude.

Q2: This looks like a "school" problem where the parts are theoretical not physical so we don't have to see how the battery reacts when you "charge" it. We just assume it looks like a voltage.

So if batteries just act like a voltage source the light is seeing +10V - 5V = 5V across it, and it will light.

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3. ### steveb Senior Member

Jul 3, 2008
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The situation in Q1 would be very dangerous in reality. Shorting batteries by any means is not a good idea.

From a theoretical point of view (which seems to be the intent of the question), you can't say much about the current (except direction) unless you model the battery source resistances. You will end up with an equivalent circuit with two voltage sources and two resistors. You can then analyze this arrangement and get your answer.

The situation in Q2 is not dangerous because the light bulb is a load that will limit current. If you assume that both batteries are identical, then no current flows and the light bulb does not light. If the batteries are not identical, their voltages will subtract and you will get a small current flow. It's likely that the current/voltage would not be enough to light up the bulb.

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4. ### PG1995 Thread Starter Active Member

Apr 15, 2011
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Thank you, ErnieM, Steve.

Sorry, I forgot to mention that the batteries in Q2 had the same voltage.

Let's say the source resistances are of the same magnitude.

I conclude that in case of Q1 there would be net current only due to 10V battery and the bulb won't light up in case of Q2.

Best wishes
PG

5. ### studiot AAC Fanatic!

Nov 9, 2007
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Last edited: Oct 31, 2011
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7. ### steveb Senior Member

Jul 3, 2008
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OK, assume that each battery has a source resistance of Rs. In theory, the current would then be I=(10-5)/(2Rs), in the direction you indicated. Note that the 5V battery still has current, but it is a charging current (storing energy). The 10V battery is supplying current and discharging energy.