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))

Please help me get on the right track

 

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.

 

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

 

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: \(P_{ower}=I^2\cdot R\)

 

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

 

I see it differently;

 

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.

 

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)