Solving a Two-Terminal Black Box Circuit

[phrygian]

Homework Statement

A two-terminal “black box” is known to contain an inductor L, a
capacitor C, and a resistor R. On connecting a 1.5 V battery, 1.5 mA
flows. When an AC voltage of 1V RMS at 60 Hz is connected, 10 mA
RMS flows. As the frequency increases at a fixed 1 V RMS, the
current reaches a maximum of over 100 A at 1 kHz. Sketch the circuit
in the black box and find values for L, C and R.

Homework Equations

The Attempt at a Solution

I can't figure out how to start this one because of the phrase "on connecting the battery". I think that must mean immediately after connecting the battery, but I don't know how the circuit could be set up if that was the case because no current can flow through the capacitor when there is zero frequency. Can someone help me see how the circuit is supposed to be set up?

 

[Andrew Mason]

What would happen if the capacitor was connected in parallel to the resistor and/or inductor?

AM

 

[phrygian]

I figured this: it can't be connected in parallel to the resistor or inductor + resistor because right after the battery is connected the potential drop across the capacitor is 0 because it hasn't charged yet, and the problem states there is current. Since there is current after the battery is connected there must be no potential across the inductor too right? Because the inductor won't allow sudden changes in current like going from 0 to 1.5 mA

 

[Andrew Mason]

I figured this: it can't be connected in parallel to the resistor or inductor + resistor because right after the battery is connected the potential drop across the capacitor is 0 because it hasn't charged yet, and the problem states there is current. Since there is current after the battery is connected there must be no potential across the inductor too right? Because the inductor won't allow sudden changes in current like going from 0 to 1.5 mA

A 0 voltage across the capacitor does not mean there is no current in the circuit. Current leads voltage in a capacitor. So there is current through the capacitor initially as charge (and voltage) is building up. When there is maximum voltage drop across the capacitor there is no current at all.

You have to assume that the current of 1.5 mA is the stable current that results from the application of 1.5 DC. The current becomes stable after a very short time so ignore initial effects.

AM