Question

The current in a coil of resistance $90\Omega$ is to be reduced by $90\%$. What value of resistance should be connected in parallel with it?

Answer 1

Step 1: Given
Resistance, $r=90\Omega$
Change in current (decrease) $=90\%$

Step 2: Formula used
$V=IR$
where
$$\begin{gathered} V=PotentialDifference \\ I=Current \\ R=Resis\tan ce \end{gathered}$$

Step 3: Calculate the value of current in both resistances

Let us assume the resistance added $=R$

$$\begin{gathered} Newcurrent,I_2=I_1-90\%of{I}_1 \\ \Rightarrow I_2=I_1-0.9I_1 \\ \Rightarrow I_2=0.1I_1 \end{gathered}$$

The current I through the resistance R $=I_1-I_2=I_1-0.1I_1=0.9I_1$
Step 4: Calculate the value of resistance to be connected in parallel

The potential difference remains the same in parallel. So, we have,

$$\begin{gathered} V_1=V_2 \\ \Rightarrow I_{2}r=IR \\ \Rightarrow 0.1I_1\times 90=0.9I_1\times R \\ \Rightarrow R=\frac{0.1\times 90}{0.9} \\ \Rightarrow R=10\Omega \end{gathered}$$

Thus, the value of resistance that should be connected in parallel with a coil of resistance of $90\Omega$ to reduce the current by 90 per cent is $10\Omega$.