## anonymous one year ago when a source is connected to a 2 ohm resistor the current is 0.5 ampere if the same source is connected to a 5 ohm resistor the current is 0.25 find the potential of the battery??

1. anonymous

@irishboy123

2. anonymous

@arindameducationusc

3. IrishBoy123

$V = IR = 2 * 0.5 \ne 5 *0.25$ post/link original question?!

4. Michele_Laino

I think that in your circuit we have to consider the internal resistance R0 of the battery, so we have the subsequent equation: $\Large E = \left( {R + {R_0}} \right)I$ where E is the electromotive force of your battery, and R is the external resistance. Using your data, we get the subsequent algebraic system of two equations: $\Large \left\{ \begin{gathered} E = \left( {2 + {R_0}} \right) \cdot 0.5 \hfill \\ E = \left( {5 + {R_0}} \right) \cdot 0.25 \hfill \\ \end{gathered} \right.$ Please solve it for E

5. IrishBoy123

@Michele_Laino cool!

6. Michele_Laino

thanks! :) @IrishBoy123

@Michele_Laino A correct and wise assumption. Most problems ignore internal resistance considering an ideal source. This case, it matters.

8. anonymous

Nice @Michele_Laino :)

9. anonymous

@PlasmaFuzer

10. anonymous

@Michele_Laino how can we get R0 because its not given???

11. Michele_Laino

If we equate, each other, both right sides of the two equations above, we get: $\Large \left( {2 + {R_0}} \right) \cdot 0.5 = \left( {5 + {R_0}} \right) \cdot 0.25$ which is an equation of first degree in R0. Please solve it for R0. @Shiraz-Khokhar