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??

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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??

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

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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
thanks! :) @IrishBoy123
@Michele_Laino A correct and wise assumption. Most problems ignore internal resistance considering an ideal source. This case, it matters.
@Michele_Laino how can we get R0 because its not given???
thanks! :) @iambatman @radar
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

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