anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@irishboy123
anonymous
  • anonymous
@arindameducationusc
IrishBoy123
  • IrishBoy123
\[V = IR = 2 * 0.5 \ne 5 *0.25\] post/link original question?!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.