The braking capacity of a circuit breaker is a measure of its ability to interrupt the flow of current in a circuit. It is typically expressed in terms of the maximum current that the circuit breaker can safely interrupt without causing damage or failure. In this case, we are given that the rated making current of the circuit breaker is 90 kA (peak) and it is rated at 22 kV.

To determine the braking capacity of the circuit breaker, we need to consider the relationship between current and voltage in a circuit. In an electrical system, the current flowing through a circuit is directly proportional to the voltage across it. This relationship is governed by Ohm's law, which states that current (I) is equal to voltage (V) divided by resistance (R), or I = V/R.

In the case of a circuit breaker, the resistance is typically very low, so we can assume it to be negligible. This means that the current flowing through the circuit breaker is primarily determined by the voltage across it. Since the circuit breaker is rated at 22 kV, this is the maximum voltage that it can safely interrupt.

To calculate the braking capacity, we can use the formula for power (P), which is equal to the product of voltage and current (P = VI). Rearranging this formula, we can solve for current (I) by dividing both sides by voltage (V), giving us I = P/V.

Given that the rated making current of the circuit breaker is 90 kA (peak) and the voltage is 22 kV, we can substitute these values into the formula to find the braking capacity:

I = 90 kA / 22 kV

Simplifying this expression, we get:

I = 4.09 kA (approx.)

Therefore, the braking capacity of the circuit breaker is approximately 4.09 kA.

However, the correct answer is given as between 1344 and 1346. It is unclear how this answer was obtained, as it does not match the calculations based on the given information. It is possible that there is additional information or a different calculation method that was used to arrive at this answer.