Question

With the switch open, roughly what must be the resistance of the resistor on the right for the current out of the battery to be the same as when the switch is closed (and the resistances of the two resistors are 20 ΩΩ and 10 ΩΩ)?

Answer 1

Question:

Assumptions

Voltage of battery = 24 V

Resistance on the right,20Ω parallel to 10 Ω resistor

Answer:

For the current out of the battery to be the same as when the switch was opened with the switch closed, the resistance on the resistor on the right must be approximately 20/3 Ω

Explanation:

We note that the switch in the assumption is on the same line as the 20 Ω resistor.

With a voltage of 24 V, and the switch closed, we have;

Total resistance =  $\frac{1}{\frac{1}{10} +\frac{1}{20} } = \frac{20}{3} \Omega$

Current out of voltage, I  = Voltage/(total resistance)

= 24 ÷ 20/3 = 24 × 3/20 = 18/5 A

Therefore, with the switch opened, we get

Resistance on the right = Initial total resistance $= \frac{20}{3} \Omega$

Therefore, with the switch opened, the resistance on the resistor on the right must be approximately equal to the resultant resistance of the two resistances in parallel.