Question

What is the capacitance such that the current through the circuit is a maximum?

Answer 1

This is what I get

Explanation:

It is already given in the question pic.
Note: It should also read as `R=4.0\ Omega`

We know that for a series `RLC` circuit,
Impedence is given as

`Z=sqrt(R^2+(X_L-X_C)^2)`

and the current and voltage are related by `V = IZ`

`=>I=V/Z`

For a fixed voltage, current will be maximum when denominator `Z` is minimum.
`=>` we need to select the capacitance in such a manner that the reactance of the capacitor cancels the reactance of the inductor and swcond term under the root sign becomes `=0`. Therefore, we get

`X_L=X_C`
`=>omegaL=1/(omegaC)`
where `omega=2pif`
`=>C=1/(omega^2L)`

Inserting given values

`C=1/((2pi60)^2(20xx10^-3))`
`=>C=3.52xx10^-4\ F`, rounded to two decimal places