Question

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??

Answer 1

@irishboy123

Answer 2

@arindameducationusc

Answer 3

\[V = IR = 2 * 0.5 \ne 5 *0.25\]

post/link original question?!

Answer 4

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

Answer 5

@Michele_Laino
cool!

Answer 6

thanks! :) @IrishBoy123

Answer 7

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

Answer 8

Nice @Michele_Laino :)

Answer 9

@PlasmaFuzer

Answer 10

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

Answer 11

thanks! :) @iambatman @radar

Answer 12

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