Solution to Question 4
The circuit is a Wheatstone bridge connected to a voltage source $V_0$. The bridge consists of fixed resistors $R_1$ and $R_2$ and a central variable rheostat. We analyze the two extreme cases given in the problem: when the rheostat resistance is very large ($R_{rh} \to \infty$) and when it is vanishingly small ($R_{rh} \to 0$).
When the resistance of the rheostat is very large, the central branch acts as an open circuit. No current flows between the top and bottom junctions. The circuit simplifies to two parallel branches:
- Branch 1: $R_1$ and $R_2$ in series. Resistance = $R_1 + R_2$.
- Branch 2: $R_2$ and $R_1$ in series. Resistance = $R_2 + R_1$.
The equivalent resistance $R_{eq1}$ is:
$$ R_{eq1} = \frac{(R_1 + R_2)(R_1 + R_2)}{(R_1 + R_2) + (R_1 + R_2)} = \frac{R_1 + R_2}{2} $$Using Ohm’s Law with the given current $I_1 = 0.75 \text{ mA}$:
$$ V_0 = I_1 R_{eq1} \implies 3.00 = (0.75 \times 10^{-3}) \left( \frac{R_1 + R_2}{2} \right) $$ $$ R_1 + R_2 = \frac{6.00}{0.75 \times 10^{-3}} = 8000 \, \Omega = 8 \, \text{k}\Omega \quad \dots(1) $$When the rheostat resistance vanishes, the top and bottom junctions are short-circuited (same potential). The circuit rearranges into two parallel pairs connected in series:
- Pair 1: $R_1$ parallel to $R_2$. Equivalent = $\frac{R_1 R_2}{R_1 + R_2}$.
- Pair 2: $R_2$ parallel to $R_1$. Equivalent = $\frac{R_1 R_2}{R_1 + R_2}$.
The total equivalent resistance $R_{eq2}$ is the sum of these pairs:
$$ R_{eq2} = \frac{R_1 R_2}{R_1 + R_2} + \frac{R_1 R_2}{R_1 + R_2} = \frac{2 R_1 R_2}{R_1 + R_2} $$Using Ohm’s Law with the given current $I_2 = 1.00 \text{ mA}$:
$$ V_0 = I_2 R_{eq2} \implies 3.00 = (1.00 \times 10^{-3}) \left( \frac{2 R_1 R_2}{R_1 + R_2} \right) $$Substituting $(R_1 + R_2) = 8000 \, \Omega$ from equation (1):
$$ 3.00 = (10^{-3}) \frac{2 R_1 R_2}{8000} $$ $$ R_1 R_2 = \frac{3.00 \times 8000}{2 \times 10^{-3}} = 12,000,000 \, \Omega^2 = 12 \, (\text{k}\Omega)^2 \quad \dots(2) $$We now have a system of equations for the sum and product of the resistances:
$$ S = R_1 + R_2 = 8 \, \text{k}\Omega $$ $$ P = R_1 R_2 = 12 \, (\text{k}\Omega)^2 $$$R_1$ and $R_2$ are the roots of the quadratic equation $x^2 – Sx + P = 0$:
$$ x^2 – 8x + 12 = 0 $$ $$ (x – 6)(x – 2) = 0 $$Thus, the values for the resistances are:
$$ x = 6 \quad \text{or} \quad x = 2 $$The values of the fixed resistances are 6 k$\Omega$ and 2 k$\Omega$.