Question
What is potential gradient? How is it measured? Explain.
✓
Answer
Consider a potentiometer consisting of a long uniform wire AB of length L and resistance R, stretched on a wooden board and connected in series with a cell of stable emf E and internal resistance r and a plug key K as shown in below figure
Let I be the current flowing through the wire when the circuit is closed.
Current through $AB , I =\frac{E}{R+r}$
Potential difference across $A B . V_{A B}=I R$
$\therefore V _{ AB }=\frac{E R}{(R+r)}$
The potential difference (the fall of potential from the high potential end) per unit length of the wire,
$\frac{V_{ AB }}{L}=\frac{E R}{(R+r) L}$
As long as $E$ and $r$ remain constant, $\frac{V_{ AB }}{L}$ will remain constant, $\frac{V_{ AB }}{L}$ is known as potential gradient along
AB and is denoted by K. Thus the potential gradient is calculated by measuring the potential difference between ends of the potentiometer wire and dividing it by the length of the wire.
Let P be any point on the wire between A and B and AP = l = length of the wire between A and P.
Then $V_{AP} = Kl$
$\therefore V_{AP} ∝ l$ as K is constant in a particular case. Thus, the potential difference across any length of the potentiometer wire is directly proportional to that length. This is the principle of the potentiometer.
Let I be the current flowing through the wire when the circuit is closed.
Current through $AB , I =\frac{E}{R+r}$
Potential difference across $A B . V_{A B}=I R$
$\therefore V _{ AB }=\frac{E R}{(R+r)}$
The potential difference (the fall of potential from the high potential end) per unit length of the wire,
$\frac{V_{ AB }}{L}=\frac{E R}{(R+r) L}$
As long as $E$ and $r$ remain constant, $\frac{V_{ AB }}{L}$ will remain constant, $\frac{V_{ AB }}{L}$ is known as potential gradient along
AB and is denoted by K. Thus the potential gradient is calculated by measuring the potential difference between ends of the potentiometer wire and dividing it by the length of the wire.
Let P be any point on the wire between A and B and AP = l = length of the wire between A and P.
Then $V_{AP} = Kl$
$\therefore V_{AP} ∝ l$ as K is constant in a particular case. Thus, the potential difference across any length of the potentiometer wire is directly proportional to that length. This is the principle of the potentiometer.
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