Question

Open circuit voltage of a source is 7.86 V and its short circuit current is 8.25 A. Find the current when an external resistance of $2.4 \Omega $ is connected to it?
A. 1.4 A
B. 1.82 A
C. 2.01 A
D. 2.4 A

Answer 1

Hint: In order to solve the above question, first we will find internal resistance of the cell by using the equation $V = IR$. Then again we will use this equation to find the new current, when resistance is connected to it, but this time we have to take total resistance to be ($R + r$) because external resistance is also included. By using this equation we can easily determine the value of current when external resistance is connected.

Complete step by step answer:
Firstly, let us understand the open-circuit voltage.
It is the voltage difference across the terminals of a cell when nothing is connected to it externally.
So, basically, it’s EMF and the current that flows through the cell when both the terminals of a cell are connected together by an ideal resistance less wire is called short circuit current.
Therefore, \[{{\text{v}}_{{\text{oc }}}} = {I_{{\text{sc}}}}{\text{r}}\] (By Ohm’s Law)
Emf, \[{{\text{v}}_{{\text{oc }}}}\]= Open circuit voltage
\[{{\text{I}}_{{\text{sc }}}}\]= Short circuit current
\[{\text{r}}\] = Internal resistance of source
$\Rightarrow 7.86 = 8.25 \times {\text{r }}$
$\Rightarrow {\text{r = 0}}{\text{.85}}\Omega $
Now, when we connect an external resistor to it then, net resistance will become \[{\text{r + R}}\].
Therefore,
\[{\text{E = I}}\left( {{\text{R + r}}} \right){\text{ }}\]
On putting the values we have
\[\Rightarrow {\text{7}}{\text{.86 = I}}\left( {{\text{2}}{\text{.4 + 0}}{\text{.85}}} \right){\text{ }}\]
\[\Rightarrow {\text{I = }}\dfrac{{7.86}}{{3.25}}\]
\[\Rightarrow {\text{I = 2}}{\text{.4A}}\]

Therefore, the current in the circuit is 2.4A. Hence, option (D) is the right answer.

Note:
Any device in an electrical circuit is said to be short-circuited when a very low resistance wire (ideally zero resistance) is connected in parallel to that device. In that situation, almost all the current flows through that ideal wire and the device is said to be short-circuited or shunted.