A \[2\]\[V\] battery, a \[15\]\[ohm\] resistor and a potentiometer of \[100\]\[cm\] length, all are connected in series, if the resistance of potentiometer wire is \[5\]\[ohm\]. The potential gradient of the potentiometer wire will be:
A. \[0.005\]\[V\]⁄\[cm\]
B. \[0.05\]\[V\]⁄\[cm\]
C. 0.02\[V\]⁄\[cm\]
D. 0.2\[V\]⁄\[cm\]
Answer
635.7k+ views
Hint: The potential gradient represents the rate of change of potential along with displacement. Also the potential gradient increases when the voltage increases and PG decreases when the length increases, as PG is directly proportional to the voltage and inversely proportional to the length.
Formula Used:
\[PG = \dfrac{V}{l}\]
\[PG\] is potential gradient, \[V\] is voltage and \[l\] is length.
Complete step by step answer:
As we know that:-
\[PG = \dfrac{V}{l}\]
Here, \[PG\] is potential gradient, \[V\] is voltage and\[l\] is length.
We also know that,
\[I = \dfrac{V}{{R'}}\] Where \[R' = R + 15\] , R is the resistance of potentiometer wire.
As it is already mentioned that the battery, \[15\,ohm\] & the potentiometer having its wire resistance of \[5\,ohm\] all are connected in series. So we can say, \[R'\] is the total resistance including the series resistance of \[15\,ohm\] & the resistance of potentiometer wire \[R = 5\,ohm\] & \[I\] is the total current.
Therefore the current I will be calculated as-
\[I = \dfrac{2}{{R + 15}}\]
As it is already given, \[R\]=\[5\] \[ohm\]
\[I = \dfrac{2}{{20}} = 0.1\]
\[\Rightarrow V = IR\]
\[\Rightarrow V = 0.1 \times 5\]
\[\Rightarrow V = 0.5\]
$\therefore$ Potential gradient = \[\dfrac{{0.5}}{{100}}\]= \[0.005\]\[V\]⁄\[cm\]
So, option A is correct.
Note:Potentiometer is a device used to measure the electromotive force, to measure internal resistance and also to measure internal resistance in a resistor or it is also defined as change in volume per unit length. The potentiometer is a simple device used to measure the electric potentials. Television, transducers, motion control, and audio equipment are some applications of potentiometer. Potentiometer is commonly used to control electronic devices. In electronics circuits, resistors are used to current flow, adjust signals, to divide voltages. While calculating the value of potential gradient, remember to take the ratio of voltage and the length.
Formula Used:
\[PG = \dfrac{V}{l}\]
\[PG\] is potential gradient, \[V\] is voltage and \[l\] is length.
Complete step by step answer:
As we know that:-
\[PG = \dfrac{V}{l}\]
Here, \[PG\] is potential gradient, \[V\] is voltage and\[l\] is length.
We also know that,
\[I = \dfrac{V}{{R'}}\] Where \[R' = R + 15\] , R is the resistance of potentiometer wire.
As it is already mentioned that the battery, \[15\,ohm\] & the potentiometer having its wire resistance of \[5\,ohm\] all are connected in series. So we can say, \[R'\] is the total resistance including the series resistance of \[15\,ohm\] & the resistance of potentiometer wire \[R = 5\,ohm\] & \[I\] is the total current.
Therefore the current I will be calculated as-
\[I = \dfrac{2}{{R + 15}}\]
As it is already given, \[R\]=\[5\] \[ohm\]
\[I = \dfrac{2}{{20}} = 0.1\]
\[\Rightarrow V = IR\]
\[\Rightarrow V = 0.1 \times 5\]
\[\Rightarrow V = 0.5\]
$\therefore$ Potential gradient = \[\dfrac{{0.5}}{{100}}\]= \[0.005\]\[V\]⁄\[cm\]
So, option A is correct.
Note:Potentiometer is a device used to measure the electromotive force, to measure internal resistance and also to measure internal resistance in a resistor or it is also defined as change in volume per unit length. The potentiometer is a simple device used to measure the electric potentials. Television, transducers, motion control, and audio equipment are some applications of potentiometer. Potentiometer is commonly used to control electronic devices. In electronics circuits, resistors are used to current flow, adjust signals, to divide voltages. While calculating the value of potential gradient, remember to take the ratio of voltage and the length.
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