How much voltage is dropped across two 5 H inductors connected in parallel with a current change of 4.5 amps per second?

To determine the voltage dropped across two 5 H inductors connected in parallel with a current change of 4.5 amps per second, we first need to use the formula for the inductive reactance given by ( V = L \frac{di}{dt} ), where ( V ) is the voltage across the inductor, ( L ) is the inductance in henries, and ( \frac{di}{dt} ) is the rate of change of current in amps per second.

Since there are two inductors connected in parallel, we first need to find the equivalent inductance. For inductors in parallel, the total inductance ( L_t ) can be calculated using the formula:

[

\frac{1}{L_t} = \frac{1}{L_1} + \frac{1}{L_2}

]

In this case, both inductors have the same inductance of 5 H:

[

\frac{1}{L_t} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}

]

[

L_t = \frac{5}{2} = 2.5 ,