A 35 k is connected in series with a resistance... - JAMB Physics 2023 Question

1. The resistances \(R_1\) and \(R_2\) in series result in a total series resistance (\(R_{\text{total, series}}\)):
   \[ R_{\text{total, series}} = R_1 + R_2 = 35 \, \text{k}\Omega + 40 \, \text{k}\Omega = 75 \, \text{k}\Omega \]

2. For resistors in parallel, the reciprocal of the total resistance (\(1/R_{\text{total, parallel}}\)) is the sum of the reciprocals of the individual resistances:
   \[ \frac{1}{R_{\text{total, parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R} \]

3. Given that the equivalent resistance is \(25 \, \text{k}\Omega\), set up the equation:
   \[ \frac{1}{25} = \frac{1}{R_{\text{total, series}}} + \frac{1}{R} \]

4. Solve for \(R\):
   \[ \frac{1}{25} = \frac{1}{75} + \frac{1}{R} \]
   \[ \frac{2}{75} = \frac{1}{R} \]
   \[ R = \frac{75}{2} = 37.5 \, \text{k}\Omega \]

Therefore, the correct option is 37.5 kΩ