A potentiometer wire of 10m length and 20 ohm resistance is connected in series with a resistance R ohm and a battery of emf 2V, negligible internal resistance. Potential gradient on the wire is 0.16 milli volt/centimeter. Then r is _ _ _ _ _ _ _ ohms
(A) 50
(B) 60
(C ) 230
(D) 46.6
Answer
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Hint: A potentiometer acts as a voltage divider. Hence current across the potentiometer wire is the ratio of the emf in the circuit to the total resistance. And the potential gradient is the ratio of the voltage to the length of the potentiometer wire. Thus by substituting the given values and rearranging we will get the resistance, r.
Complete step by step answer:
Resistance of the wire, R=20 ohm
Length of potentiometer wire, L=10m
emf=2V
Potential gradient=0.16 mV/cm
Thus current across the wire, $I=\dfrac{E}{R+r}$
Potential gradient$=\dfrac{V}{L}=\dfrac{\dfrac{ER}{R+r}}{L}=0.016$
$\begin{align}
& \dfrac{ER}{R+r}=0.016\times L \\
& \\
\end{align}$
Substituting the values we get,
$\dfrac{2\times 20}{20+r}=0.016\times 10$
$\Rightarrow \dfrac{40}{20+r}=0.16$
$40=0.16\times \left( 20+r \right)$
$40=3.2+0.16r$
$40-3.2=0.16r$
$36.8=0.16r$
$r=\dfrac{36.8}{0.16}$
$\therefore r=230ohm$
So, the correct answer is “Option C”.
Additional Information: Copper is not used to make a potentiometer wire. The low temperature coefficient wires are used for making potentiometer wire. As the length of the potential wire is increased the sensitivity of the potentiometer wire can also be increased. It is the smallest value of potential difference that can be measured using a potentiometer. The emf of a given cell can be measured using a potentiometer. Also internal resistance can be calculated using a potentiometer. The emf of two cells can be compared using this. But because of the inconvenience it is used less. Compared to a voltmeter potentiometer can be used to find the actual value of emf of a cell.
Note: The potentiometer works for a constant current. The potential difference is directly proportional to length of the wire. That is the principle of the potentiometer. The emf of a given cell can be measured using a potentiometer. Also internal resistance can be calculated using a potentiometer. The emf of two cells can be compared using this.
Complete step by step answer:
Resistance of the wire, R=20 ohm
Length of potentiometer wire, L=10m
emf=2V
Potential gradient=0.16 mV/cm
Thus current across the wire, $I=\dfrac{E}{R+r}$
Potential gradient$=\dfrac{V}{L}=\dfrac{\dfrac{ER}{R+r}}{L}=0.016$
$\begin{align}
& \dfrac{ER}{R+r}=0.016\times L \\
& \\
\end{align}$
Substituting the values we get,
$\dfrac{2\times 20}{20+r}=0.016\times 10$
$\Rightarrow \dfrac{40}{20+r}=0.16$
$40=0.16\times \left( 20+r \right)$
$40=3.2+0.16r$
$40-3.2=0.16r$
$36.8=0.16r$
$r=\dfrac{36.8}{0.16}$
$\therefore r=230ohm$
So, the correct answer is “Option C”.
Additional Information: Copper is not used to make a potentiometer wire. The low temperature coefficient wires are used for making potentiometer wire. As the length of the potential wire is increased the sensitivity of the potentiometer wire can also be increased. It is the smallest value of potential difference that can be measured using a potentiometer. The emf of a given cell can be measured using a potentiometer. Also internal resistance can be calculated using a potentiometer. The emf of two cells can be compared using this. But because of the inconvenience it is used less. Compared to a voltmeter potentiometer can be used to find the actual value of emf of a cell.
Note: The potentiometer works for a constant current. The potential difference is directly proportional to length of the wire. That is the principle of the potentiometer. The emf of a given cell can be measured using a potentiometer. Also internal resistance can be calculated using a potentiometer. The emf of two cells can be compared using this.
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