Question

If the current flowing in a coil of resistance $90\,\Omega$ is to be reduced to 90%, then what value of resistance should be connected in parallel with it?
A. $90\,\Omega$
B. $9\,\Omega$
C. $1000\,\Omega$
D. $10\,\Omega$

Answer 1

We know that the voltage across the parallel circuit remains constant. Therefore, the voltage across the coil and voltage across the parallel resistor is the same. Use Ohm’s law for voltage to solve this question.

Formula used:
$V = IR$
Here, V is the voltage, I is the current, and R is the resistance.

Complete step by step answer:
We know that the voltage in the parallel circuit remains constant and the current divides. In the series circuit, the current remains constant and voltage drops at each component of the circuit.
According to Ohm’s law, the voltage in the coil is, $V = IR$, where I is the current and R is the resistance.
As per the given question, when we connect the resistance parallel to the coil, the current in the coil reduces to 90% of the initial current as the portion of the current now started flowing through the resistance we connected in parallel. Therefore, after we connect the resistance parallel to the coil, the current in the coil will become 10% of the initial current and remaining 90% current will flow through the parallel resistor.
Since the voltage across the coil and the voltage across the new resistor is the same, we can write,
$\left( {0.1} \right){I_{initial}}R = \left( {0.9{I_{initial}}} \right){R_{new}}$
$\Rightarrow {R_{new}} = \dfrac{R}{9}$
Substitute $90\,\Omega$ for R in the above equation.
${R_{new}} = \dfrac{{90}}{9}$
$\Rightarrow {R_{new}} = 10\,\Omega$
Therefore, resistance of $10\,\Omega$ should be connected parallel to the coil.

So, the correct answer is “Option D”.

Note:
We were asked to determine the value of parallel resistance to reduce the current in the coil by 90%.
Therefore, the final current in the coil must be 10% of the initial value and not 90% of the initial value.