# calculating current with parallel batteries

Discussion in 'Homework Help' started by msedtal, Nov 11, 2009.

1. ### msedtal Thread Starter New Member

Nov 11, 2009
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0
i am in a physics 2 course, and i am having trouble calculating current with two batteries connected in parrallel. I am at a complete loss of where to start. The problem is probably simple to most of you.

From the problem i am thinking that i should use I=V/R ; but am unsure if V=9+9; and am also unsure about R, i am thinking it should be R=1+ (1/(1/2)+(1/3))

2. ### JDT Well-Known Member

Feb 12, 2009
658
87
Start by noticing that the circuit can be re-drawn. See Diagram.

From circuit B, work out the current through R3. Then work back.

There is a more mathematical way to do it but because both battery voltages are the same this is the easy way. Only Ohms law needed.

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3. ### msedtal Thread Starter New Member

Nov 11, 2009
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thanks for the help, it put me on the right track and i got the questions correct.

4. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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It appears that the top 1Ω Resistor is the load, the other two resistors represent the internal resistance of the batteries (which goes up as a battery goes dead).

Help for the bonus point: $P_{ower}=I^2\cdot R$

5. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,182
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Number one rule of Electronics.....never make anything more complicated than it really is!

In a PARALLEL circuit, the voltage is the same across all components. The battry will appear as a single voltage source. Ohm's law does the rest.

eric

6. ### GetDeviceInfo AAC Fanatic!

Jun 7, 2009
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I see it differently;

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7. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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Are you sure about the polarity of the batteries in the OP, to be parallel?

They appear to be connected in series, as the post above shows.

8. ### Quintilis_Telescope New Member

Oct 13, 2009
10
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Loop 1 = 3Ω*I1 + 1Ω(I1-I2) = 9V
Loop 2 = 2Ω*I2 + 1Ω(I2-I1) = 9V

I1 = 3.2727A
I2 = 4.0909A
I3 = 3.2727A-4.0909A = -818mA (flowing up)

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