
How many \[176\Omega \] resistors are required to carry \[5\,A\] on a \[220V\] line.
Answer
508.2k+ views
Hint: Study about the relation between the current and the voltage of a circuit. The relation between current and voltage is given by the Ohm’s law which states that the voltage of a circuit if other physical quantities are kept constant is directly proportional to the current flowing through the circuit
Formula used: The mathematical expression of the Ohm’s law is given by,
\[V = IR\]
where, \[V\] is the potential difference of the circuit element,\[R\] is the resistance of the circuit element and \[I\] is the current through it.
The equivalent resistance in a parallel circuit is given by,
\[\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ....\]
where, \[{R_{eq}}\] is the equivalent resistance of the circuit and \[{R_1}\] ,\[{R_2}\]… are the resistances connected in parallel.
Complete step by step answer:
We have given here to find the number of \[176\Omega \] resistors needed to carry \[5A\] on a \[220V\] line. Let, the number of resistors needed is \[x\]. Hence, the equivalent resistance of the \[x\] number of resistors will be when connected in parallel,
\[\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{176}} + \dfrac{1}{{176}} + \dfrac{1}{{176}} + \dfrac{1}{{176}}....upto\,x\,terms\]
\[\Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{x}{{176}}\]
\[\Rightarrow {R_{eq}} = \dfrac{{176}}{x}\]
Now, from Ohm’s law we have where, is the potential difference of the circuit element, is the resistance of the circuit element and is the current through it.
Putting the value of , and \[V = IR\] we have,
\[220 = \dfrac{{176}}{x} \times 5\]
\[\Rightarrow x = \dfrac{{176 \times 5}}{{220}}\]
\[\therefore x = 4\]
Hence, we need a total of \[4\] resistance to carry \[5\,A\] on a \[220\,V\] line.
Note:The number of required resistors can be decreased by connecting the resistors in series easily. The resistance between two points is nothing but the measure of the repulsive force acting when current flows. This repulsive force to the current is actually due to the attraction by the nucleus in the material which slows down the flowing electrons by the attraction.
Formula used: The mathematical expression of the Ohm’s law is given by,
\[V = IR\]
where, \[V\] is the potential difference of the circuit element,\[R\] is the resistance of the circuit element and \[I\] is the current through it.
The equivalent resistance in a parallel circuit is given by,
\[\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ....\]
where, \[{R_{eq}}\] is the equivalent resistance of the circuit and \[{R_1}\] ,\[{R_2}\]… are the resistances connected in parallel.
Complete step by step answer:
We have given here to find the number of \[176\Omega \] resistors needed to carry \[5A\] on a \[220V\] line. Let, the number of resistors needed is \[x\]. Hence, the equivalent resistance of the \[x\] number of resistors will be when connected in parallel,
\[\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{176}} + \dfrac{1}{{176}} + \dfrac{1}{{176}} + \dfrac{1}{{176}}....upto\,x\,terms\]
\[\Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{x}{{176}}\]
\[\Rightarrow {R_{eq}} = \dfrac{{176}}{x}\]
Now, from Ohm’s law we have where, is the potential difference of the circuit element, is the resistance of the circuit element and is the current through it.
Putting the value of , and \[V = IR\] we have,
\[220 = \dfrac{{176}}{x} \times 5\]
\[\Rightarrow x = \dfrac{{176 \times 5}}{{220}}\]
\[\therefore x = 4\]
Hence, we need a total of \[4\] resistance to carry \[5\,A\] on a \[220\,V\] line.
Note:The number of required resistors can be decreased by connecting the resistors in series easily. The resistance between two points is nothing but the measure of the repulsive force acting when current flows. This repulsive force to the current is actually due to the attraction by the nucleus in the material which slows down the flowing electrons by the attraction.
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