
An electric heater has resistance $150\,\Omega $ and can bear a maximum current of $1\,A$. If the heater is to be used on $220\,V$ mains, the least resistance required in the circuit will be:
(A) $70\,\Omega $
(B) $5\,\Omega $
(C) $2.5\,\Omega $
(D) $1.4\,\Omega $
Answer
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Hint:In this problem, the resistance, current and voltage is given. By using the relation between the voltage, current and resistance the least resistance required can be determined. The equation which shows the relation between the voltage, current and resistance is Ohm’s law. By using this law the resistance can be determined.
Formulae Used:
Ohm’s law states that the voltage across any circuit is equal to the product of the current and the resistance.
$V = IR$
Where, $V$ is the voltage, $I$ is the current and $R$ is the resistance.
Complete step-by-step solution:
Given that,
The resistance of the heater, ${R_1} = 150\,\Omega $
The maximum current in the heater, $I = 1\,A$
The voltage of the heater, $V = 220\,V$
By ohm’s law,
$V = IR\,.................\left( 1 \right)$
Here, the least resistance is added additionally, so the given resistance is taken as ${R_1}$ and the resistance added is taken as ${R_2}$, then the above equation (1) is written as,
$V = I \times \left( {{R_1} + {R_2}} \right)\,..................\left( 2 \right)$
By substituting the voltage, current and the resistance value in the above equation. Then
$220 = 1 \times \left( {150 + {R_2}} \right)$
On multiplying the terms in RHS, then the above equation is written as,
$220 = 150 + {R_2}$
We have to find ${R_2}$, so keep the term ${R_2}$ in one side and other terms in other side, then the above equation is written as,
${R_2} = 220 - 150$
On subtracting, the above equation is written as,
${R_2} = 70\,\Omega $
Thus, the above equation shows the least resistance required in the circuit.
Hence, the option (A) is the correct answer.
Note:- In this problem, the least resistance is added, so we take two resistance, one resistance value is given in the problem and the other resistance is taken as ${R_2}$ and this resistance is determined by using Ohm's law.
Formulae Used:
Ohm’s law states that the voltage across any circuit is equal to the product of the current and the resistance.
$V = IR$
Where, $V$ is the voltage, $I$ is the current and $R$ is the resistance.
Complete step-by-step solution:
Given that,
The resistance of the heater, ${R_1} = 150\,\Omega $
The maximum current in the heater, $I = 1\,A$
The voltage of the heater, $V = 220\,V$
By ohm’s law,
$V = IR\,.................\left( 1 \right)$
Here, the least resistance is added additionally, so the given resistance is taken as ${R_1}$ and the resistance added is taken as ${R_2}$, then the above equation (1) is written as,
$V = I \times \left( {{R_1} + {R_2}} \right)\,..................\left( 2 \right)$
By substituting the voltage, current and the resistance value in the above equation. Then
$220 = 1 \times \left( {150 + {R_2}} \right)$
On multiplying the terms in RHS, then the above equation is written as,
$220 = 150 + {R_2}$
We have to find ${R_2}$, so keep the term ${R_2}$ in one side and other terms in other side, then the above equation is written as,
${R_2} = 220 - 150$
On subtracting, the above equation is written as,
${R_2} = 70\,\Omega $
Thus, the above equation shows the least resistance required in the circuit.
Hence, the option (A) is the correct answer.
Note:- In this problem, the least resistance is added, so we take two resistance, one resistance value is given in the problem and the other resistance is taken as ${R_2}$ and this resistance is determined by using Ohm's law.
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