Through the electrolytic bath connected to the network, a charge of 3.6kC

Through the electrolytic bath connected to the network, a charge of 3.6kC passes in 6 minutes. The mains voltage is 9V. Determine the resistance of the bath.

These tasks: t (the duration of the passage of the charge through the electrolytic bath) = 6 minutes (in SI t = 360 s); q (charge passed through the bath) = 3.6 kC (3.6 * 10 ^ 3 C); U (mains voltage) = 9 V.

The resistance of a given electrolytic bath can be calculated by the formula: R = U / I = U / (q / t) = U * t / q.

Let’s calculate: R = 9 * 360 / (3.6 * 10 ^ 3) = 0.9 Ohm.

Answer: The resistance of this electrolytic bath is 0.9 ohm.