
Write any two factors on which internal resistance of a cell depends. The reading on a high resistance voltmeter, when a cell is connected across it, is 2V. When the terminals of the cell are also connected to a resistance of 3Ω the voltmeter reading drops to 1.5V. Find the internal resistance of the cell.
Answer
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Hint:Inside a cell, the flow of current also feels some opposition and this opposition is termed as the internal resistance of the cell. But this internal resistance is very minute as compared to the external resistance. There are many methods to determine this internal resistance of the cell.
Formula used:
We can make use of the following formula to find out the internal resistance of the cell $r=(\dfrac{E}{V}-1)R$
where E is the emf and v is the potential difference and R is the external resistance.
Complete answer:
The internal resistance of the cell depends on many factors and some of them are listed below:
-It is directly proportional to the distance between the electrodes.
-It is inversely proportional to the area of the electrodes.
-It increases with the concentration of the solution that is the electrolyte.
Now given the reading on a high resistance voltmeter, when the cell is connected is 2V and when the terminals of the cell are also connected to 3Ω the voltmeter reading drops to 1.5V. It means emf is 2V and the potential difference is 1.5 V. Thus,
$r=(\dfrac{E}{V}-1)R$
$\Rightarrow r=(\dfrac{2}{1.5}-1)3 \\
\Rightarrow r=(\dfrac{2-1.5}{1.5})\times 3 \\
\Rightarrow r=(\dfrac{0.5}{1.5})\times 3 \\
\therefore r=1\Omega $
So, the internal resistance comes out to be \[1\Omega \].
Note:While solving such problems we need to keep in mind that the emf is the maximum potential difference that a cell can deliver when it is not connected to any circuit. All the values must be taken in standard SI units. The electromotive force or emf is always greater than the potential difference across the same cell. Current is always taken in units of ampere.
Formula used:
We can make use of the following formula to find out the internal resistance of the cell $r=(\dfrac{E}{V}-1)R$
where E is the emf and v is the potential difference and R is the external resistance.
Complete answer:
The internal resistance of the cell depends on many factors and some of them are listed below:
-It is directly proportional to the distance between the electrodes.
-It is inversely proportional to the area of the electrodes.
-It increases with the concentration of the solution that is the electrolyte.
Now given the reading on a high resistance voltmeter, when the cell is connected is 2V and when the terminals of the cell are also connected to 3Ω the voltmeter reading drops to 1.5V. It means emf is 2V and the potential difference is 1.5 V. Thus,
$r=(\dfrac{E}{V}-1)R$
$\Rightarrow r=(\dfrac{2}{1.5}-1)3 \\
\Rightarrow r=(\dfrac{2-1.5}{1.5})\times 3 \\
\Rightarrow r=(\dfrac{0.5}{1.5})\times 3 \\
\therefore r=1\Omega $
So, the internal resistance comes out to be \[1\Omega \].
Note:While solving such problems we need to keep in mind that the emf is the maximum potential difference that a cell can deliver when it is not connected to any circuit. All the values must be taken in standard SI units. The electromotive force or emf is always greater than the potential difference across the same cell. Current is always taken in units of ampere.
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