Calculating Battery Current in a Parallel-Series Circuit with Resistances

[kevindesai777]

Homework Statement

Three resistances of 4,6,12 ohms are connected in parallel and the combination is connected in series with a 4V battery with internal resistance 2 ohms The battery current is what

Homework Equations

No relevant question.


[b]3. The Attempt at a Solution [/b]
I don't know

 

[willem2]

replace parallel or series combinations of resistances by their equivalent until you have only 1 resistance left.

 

[dreamspace]

I = $$\frac{E}{R}$$ = $$\frac{4V}{Ri+(R1||R2||R3)}$$ = $$\frac{4V}{2 ohm + (\frac{1}{4 ohm}+\frac{1}{6 ohm}+\frac{1}{12 ohm})}$$ = $$\frac{4V}{2 ohm}$$ = 2A

 

[willem2]

What happened to this? $$(\frac{1}{4 ohm}+\frac{1}{6 ohm}+\frac{1}{12 ohm})$$

 

[kevindesai777]

I = $$\frac{E}{R}$$ = $$\frac{4V}{Ri+(R1||R2||R3)}$$ = $$\frac{4V}{2 ohm + (\frac{1}{4 ohm}+\frac{1}{6 ohm}+\frac{1}{12 ohm})}$$ = $$\frac{4V}{2 ohm}$$ = 2A

thank u

 

[willem2]

I didn't notice it the first time but (R1||R2||R3) =

$$1 / (\frac{1}{4 ohm}+\frac{1}{6 ohm}+\frac{1}{12 ohm})$$

The answer is still right.