Answer to Question #147056 in Electric Circuits for Sridhar

Question

When a LR circuit with L=3mH is connected in series to an AC source of rms voltage 30V a maximum RMS current of 6A is observed at frequency (500/π)Hz. If LR circuit is now connected to a battery of emf 21V And internal resistance of 3ohm the current will be
Ans 3A

Answer 1

Let the RMS current be $I$ $I^2 = \frac{V^2}{R^2+(wL)^2}$

$w=2(pi)f$

$I=6A$

$f=\frac{500}{pi}$

Where $pi=\frac{22}{7}$

Solving the above equations we get R= 4 ohm.

When the LR circuit is connected to battery of emf 21V and internal resistance 3 ohm then new current is i .

$i =\frac{21}{4+3}$

$i= 3A$