91Ó°ÊÓ

The change in current through battery is 1.25% .

You have an array of resistor in parallel. You have to find the current with n number of resistors and the current when one more resistor is added. You can find the total number of resistors in the array by using the given value of percentage change in the current through battery.

Formula:

Equivalent resistance when n resistor in parallel:

$$\begin{gathered} R_{eq}=R+\frac{R}{n} \\ =\frac{R\left(n+1\right)}{n} \end{gathered}$$

Equivalent resistance when n + 1 resistor in parallel:

$$\begin{gathered} R_{eq}=R+\frac{R}{n+1} \\ =\frac{R\left(n+2\right)}{n+1} \end{gathered}$$

Current in block:

$i=\frac{V}{R}$

Here, i is the current, R is the resistor, V is the voltage, n is the number of array, and R_eq is the equivalent resistor.

The current in the battery in the 1^st case:

$$\begin{gathered} i_n=\frac{V_{battery}}{R_{eq}} \\ =\left(\frac{n}{n+1}\right)\frac{V_{battery}}{R} \end{gathered}$$

The current in the battery in the 2^nd case:

$$\begin{gathered} i_{n+1}=\frac{V_{battery}}{R_{eq}} \\ =\left(\frac{n+1}{n+2}\right)\frac{V_{battery}}{R} \end{gathered}$$

Since the relative increase in current is 1.25%, you can write

$$\begin{gathered} \frac{i_{n+1}-i_n}{i_n}=\frac{1.25}{100} \\ =0.0125 \\ \frac{\left[\left(\frac{n+1}{n+2}\right)\frac{V_{battery}}{R}\right]-\left[\left(\frac{n}{n+1}\right)\frac{V_{battery}}{R}\right]}{\frac{n}{n+1}\frac{V_{battery}}{R}}=0.0125 \\ \frac{\left(\frac{n+1}{n+2}\right)-\left(\frac{n}{n+1}\right)}{\frac{n}{n+1}}=0.0125 \\ \frac{\left(\frac{n+1}{n+2}\right)}{\frac{n}{n+1}}-1=0.0125\backslash \end{gathered}$$

$$\begin{gathered} \frac{\left(\frac{n+1}{n+2}\right)}{\frac{n}{n+1}}=1.0125 \\ \left(\frac{n+1}{n+2}\right)×\frac{n+1}{n}=1.0125 \\ \frac{(n^2+2n+1)}{n^2+2n}=1.0125 \\ (n^2+2n+1)=1.0125n^2+2.025n \\ 0.0125n^2+0.025n-1=0 \end{gathered}$$

Divide both sides by 0.125 and you get,

$n^2+2n-80=0$

After solving the quadratic equation, you get

$$\begin{gathered} n=\frac{-2Â\pm \sqrt{{\left(-2\right)}^2-4\left(1\right)\left(-80\right)}}{2\left(1\right)} \\ =\frac{-2Â\pm \sqrt{4+320}}{2} \\ =\frac{-2Â\pm 18}{2} \end{gathered}$$

As array is positive.

$$\begin{gathered} n=\frac{-2+18}{2} \\ =\frac{16}{2} \\ =8 \end{gathered}$$

Hence, the required value of n is 8 .