
A potentiometer wire of length \[L\] and a resistance \[r\] are connected in series with a battery of e.m.f. \[{E_0}\] and a resistance \[{r_1}\]. An unknown e.m.f. \[E\] is balanced at a length \[l\] of the potentiometer wire. The e.m.f. \[E\] will be given by:
A. \[\dfrac{{L{E_0}r}}{{\left( {r + {r_1}} \right)1}}\]
B. \[\dfrac{{L{E_0}r}}{{l{r_1}}}\]
C. \[\dfrac{{{E_0}r}}{{\left( {r + {r_1}} \right)}}\dfrac{l}{L}\]
D. \[\dfrac{{{E_0}l}}{L}\]
Answer
568.5k+ views
Hint:First of all, we will use the Ohm’s law to find the current flowing. After that we will find the potential difference across the potentiometer followed by the potential gradient. We know e.m.f. is given by the product of potential gradient and length. We will manipulate accordingly to find the result.
Formula used:
Using Ohm’s formula,
\[I = \dfrac{V}{{{R_{eq}}}}\] …… (1)
Where,
\[I\] is current
\[V\] is voltage
\[R\] is resistance.
Complete step by step answer:
According to question the current will be,
By using Ohm’s law,
$I = \dfrac{V}{{{R_{eq}}}} \\
\Rightarrow I = \dfrac{{{E_0}}}{{r + {r_1}}} \\$
The potential difference across potentiometer is given by:
\[I \times r = \dfrac{{{E_0}r}}{{\left( {r + {r_1}} \right)}}\].
The potential gradient will be
$\dfrac{V}{L} = \dfrac{{{E_0}r}}{{\left( {r + {r_1}} \right)}} \times \dfrac{1}{L} \\
\Rightarrow \dfrac{V}{L} = k \\$
We know that,
\[{\text{EMF}} = k \times l\]
Where \[l\]is balancing length,
So,
The emf will be,
${\text{EMF}} = \left( {\dfrac{{{E_0}r}}{{r + {r_1}}}} \right)\dfrac{1}{L} \\
\therefore {\text{EMF}} = \dfrac{{{E_0}r}}{{\left( {r + {r_1}} \right)}}\dfrac{1}{L} \\$
Hence, the required answer is \[\dfrac{{{E_0}r}}{{\left( {r + {r_1}} \right)}}\dfrac{l}{L}\].
The correct option is C.
Additional information:
Potentiometer: A potentiometer is a three-terminal resistor forming an adjustable voltage divider with a sliding or rotating contact. It functions as a variable resistor, or rheostat, if only two terminals are used, one end and the wiper. The potentiometer measuring instrument is basically a voltage divider used to measure electrical potential (voltage); the component applies the same concept, hence its name.
Note: Remember that the potentiometer measuring instrument is basically a voltage divider used to measure electrical potential (voltage); the component applies the same concept, hence its name. Potentiometers are typically used on audio equipment to control electrical devices such as volume controls.
Formula used:
Using Ohm’s formula,
\[I = \dfrac{V}{{{R_{eq}}}}\] …… (1)
Where,
\[I\] is current
\[V\] is voltage
\[R\] is resistance.
Complete step by step answer:
According to question the current will be,
By using Ohm’s law,
$I = \dfrac{V}{{{R_{eq}}}} \\
\Rightarrow I = \dfrac{{{E_0}}}{{r + {r_1}}} \\$
The potential difference across potentiometer is given by:
\[I \times r = \dfrac{{{E_0}r}}{{\left( {r + {r_1}} \right)}}\].
The potential gradient will be
$\dfrac{V}{L} = \dfrac{{{E_0}r}}{{\left( {r + {r_1}} \right)}} \times \dfrac{1}{L} \\
\Rightarrow \dfrac{V}{L} = k \\$
We know that,
\[{\text{EMF}} = k \times l\]
Where \[l\]is balancing length,
So,
The emf will be,
${\text{EMF}} = \left( {\dfrac{{{E_0}r}}{{r + {r_1}}}} \right)\dfrac{1}{L} \\
\therefore {\text{EMF}} = \dfrac{{{E_0}r}}{{\left( {r + {r_1}} \right)}}\dfrac{1}{L} \\$
Hence, the required answer is \[\dfrac{{{E_0}r}}{{\left( {r + {r_1}} \right)}}\dfrac{l}{L}\].
The correct option is C.
Additional information:
Potentiometer: A potentiometer is a three-terminal resistor forming an adjustable voltage divider with a sliding or rotating contact. It functions as a variable resistor, or rheostat, if only two terminals are used, one end and the wiper. The potentiometer measuring instrument is basically a voltage divider used to measure electrical potential (voltage); the component applies the same concept, hence its name.
Note: Remember that the potentiometer measuring instrument is basically a voltage divider used to measure electrical potential (voltage); the component applies the same concept, hence its name. Potentiometers are typically used on audio equipment to control electrical devices such as volume controls.
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