Answer to Question #93303 in Electric Circuits for Hating Isah Adam

Question

An a.c of voltage 50Hz is connected to an inductor of 2H and neglect table resistant. The current r.m.s flow in the coil when the frequent of voltage is changed to 400Have keeping the magnitude on the voltage thesame, the current flowing now is?

Answer 1

The text of the problem is a bit unclear, but thanks to the universal language understood by anyone who knows math - the language of formulas. At 50 Hz the reactance of the coil is:

$$X_{L,50}=2\pi f_{50}L.$$

The current while the frequency is 50 Hz is:

$$I_{50}=\frac{V}{X_{L,50}},$$

or if we express the voltage:

$$V=I_{50}\cdot X_{L,50}.$$

Then, when the frequency is changed to 400 Hz, the current will be

$$I_{400}=\frac{V}{X_{L,400}},$$

substituting the previous expression for voltage we can calculate the current at 400 Hz:

$$I_{400}=\frac{I_{50}\cdot X_{L,50}}{X_{L,400}}=I_{50}\frac{f_{50}}{f_{400}}.$$