Answer
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Hint: This question requires the substitution of values in the basic formulae of voltage. Using the values of current and resistance given, we will be able to find the voltage drop across the wire.
Formulae used:
Current: ${V_d} = IR$
Where $R$ is the resistances in the circuit and is expressed in Ohms $(\Omega )$, $I$ is the current in the circuit and is expressed in Amperes $(A)$ and ${V_d}$ is the potential difference across the capacitor and is expressed in Coulombs $(C)$.
Complete step by step answer:
It is said in the question that a wire has $0.1A$ current flowing in it while a resistance $120m\Omega /m$ inhibits the same. We are to determine the potential difference across $1m$ of the same wire. We extract the following data from the question:
$R = 120m\Omega /m$, $I = 0.1A$, Length of wire $ = 1m$
Resistance for the given wire is,
\[R \times length\,of\,wire = 120m\Omega /m\,\, \times \,1m = 120m\Omega \]
Converting the unit of resistance into SI unit $120 \times {10^{ - 3}}\Omega = 0.120\Omega $
Potential difference is basically the difference between electric potential at two points. Famously it is also directly calculated using Ohm’s Law.
It is observed that the amount of current flowing in the circuit, that is $0.1A$, is quite small. Also, the resistance is also quite inconsequential, $0.12\Omega $. Similarly, the potential across the wire due to its internal resistance will be small as well as it is directly proportional to the current and resistance both.
Using Ohm’s Law we get ${V_d} = IR$.
Substituting the values of resistance $(R)$ and current $(I)$ we get,
${V_d} = IR = 0.1A\, \times \,0.12\Omega = 0.012A\,\Omega $
But $A\,\Omega = V$.
$\therefore{V_d} = 0.012V$.
Thus, the potential difference across $1m$ of wire is $0.012V$.
Note: The resistance value in the given question is per meter length of the wire wise. it basically means that the resistance of the entire length of the wire is as such and we need to multiply with it the length of whatever segment of it is provided in the question in order to get that particular segment’s resistance. Potential difference is basically the difference between the voltages at two given points. But it can also be directly calculated using current and resistance values. Potential difference is basically the difference between the voltages at two given points. But it can also be directly calculated using current and resistance values.
Formulae used:
Current: ${V_d} = IR$
Where $R$ is the resistances in the circuit and is expressed in Ohms $(\Omega )$, $I$ is the current in the circuit and is expressed in Amperes $(A)$ and ${V_d}$ is the potential difference across the capacitor and is expressed in Coulombs $(C)$.
Complete step by step answer:
It is said in the question that a wire has $0.1A$ current flowing in it while a resistance $120m\Omega /m$ inhibits the same. We are to determine the potential difference across $1m$ of the same wire. We extract the following data from the question:
$R = 120m\Omega /m$, $I = 0.1A$, Length of wire $ = 1m$
Resistance for the given wire is,
\[R \times length\,of\,wire = 120m\Omega /m\,\, \times \,1m = 120m\Omega \]
Converting the unit of resistance into SI unit $120 \times {10^{ - 3}}\Omega = 0.120\Omega $
Potential difference is basically the difference between electric potential at two points. Famously it is also directly calculated using Ohm’s Law.
It is observed that the amount of current flowing in the circuit, that is $0.1A$, is quite small. Also, the resistance is also quite inconsequential, $0.12\Omega $. Similarly, the potential across the wire due to its internal resistance will be small as well as it is directly proportional to the current and resistance both.
Using Ohm’s Law we get ${V_d} = IR$.
Substituting the values of resistance $(R)$ and current $(I)$ we get,
${V_d} = IR = 0.1A\, \times \,0.12\Omega = 0.012A\,\Omega $
But $A\,\Omega = V$.
$\therefore{V_d} = 0.012V$.
Thus, the potential difference across $1m$ of wire is $0.012V$.
Note: The resistance value in the given question is per meter length of the wire wise. it basically means that the resistance of the entire length of the wire is as such and we need to multiply with it the length of whatever segment of it is provided in the question in order to get that particular segment’s resistance. Potential difference is basically the difference between the voltages at two given points. But it can also be directly calculated using current and resistance values. Potential difference is basically the difference between the voltages at two given points. But it can also be directly calculated using current and resistance values.
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