Solution to Question 9

Step 1: Set up the equations

Case 1: Only R is connected.
$$ I = \frac{\varepsilon}{R + r} = \frac{\varepsilon}{100 + r} \quad \dots(1) $$

Case 2: X is in series with R.
The new current is $3I/4$. Total resistance is $R + X + r$. $$ \frac{3I}{4} = \frac{\varepsilon}{100 + X + r} \quad \dots(2) $$

Case 3: X is in parallel with R.
The new current is $6I/5$. The equivalent external resistance is $\frac{RX}{R+X}$. $$ \frac{6I}{5} = \frac{\varepsilon}{\frac{100X}{100+X} + r} \quad \dots(3) $$

Step 2: Solve the system of equations

Divide (1) by (2):

$$ \frac{I}{3I/4} = \frac{\frac{\varepsilon}{100+r}}{\frac{\varepsilon}{100+X+r}} \implies \frac{4}{3} = \frac{100+X+r}{100+r} $$ $$ 4(100+r) = 3(100+X+r) $$ $$ 400 + 4r = 300 + 3X + 3r $$ $$ r = 3X – 100 \quad \dots(4) $$

Divide (3) by (1):

$$ \frac{6I/5}{I} = \frac{\frac{\varepsilon}{\frac{100X}{100+X} + r}}{\frac{\varepsilon}{100+r}} \implies \frac{6}{5} = \frac{100+r}{\frac{100X}{100+X} + r} $$ $$ 6\left( \frac{100X}{100+X} + r \right) = 5(100+r) $$ $$ \frac{600X}{100+X} + 6r = 500 + 5r $$ $$ r = 500 – \frac{600X}{100+X} \quad \dots(5) $$

Step 3: Equate r and solve for X

Equating (4) and (5):

$$ 3X – 100 = 500 – \frac{600X}{100+X} $$ $$ 3X – 600 = -\frac{600X}{100+X} $$

Divide by 3:

$$ X – 200 = -\frac{200X}{100+X} $$ $$ (X – 200)(X + 100) = -200X $$ $$ X^2 + 100X – 200X – 20000 = -200X $$ $$ X^2 – 100X – 20000 = -200X $$ $$ X^2 + 100X – 20000 = 0 $$

Solving the quadratic equation:

$$ X = \frac{-100 \pm \sqrt{100^2 – 4(1)(-20000)}}{2} $$ $$ X = \frac{-100 \pm \sqrt{10000 + 80000}}{2} = \frac{-100 \pm 300}{2} $$

Since resistance must be positive, take the positive root:

$$ X = \frac{200}{2} = 100\,\Omega $$