Question

this problem we consider the resonance curve for a circuit electrical components ( l ), ( r ), and ( c ) and resonant equency ( omega_0 ).
for a given value of ( c ), the resonance frequency is inversely proportional to the square root of ( l ). similarly, for a given value of ( l ), the resonance frequency is inversely proportional to the square root of ( c ).
part b
for given values of ( r ) and ( c ), if you double the value of ( l ), how does the new rms current at resonance ( i_{\text{rms}} ) differ from its original value? assume that the voltage amplitude of the ac source is the same
view available hint(s)
( i_{\text{rms}} ) is twice as great.
( i_{\text{rms}} ) is half as great.
( i_{\text{rms}} ) is ( 1/sqrt{2} ) times as great.
( i_{\text{rms}} ) is unchanged.

Explanation:

Step1: Recall the formula for \(I_{rms}\) at resonance

At resonance, the impedance \(Z = R\) (since \(X_L=X_C\)). The formula for \(I_{rms}\) is \(I_{rms}=\frac{V_{rms}}{Z}\). And \(V_{rms}=\frac{V_{max}}{\sqrt{2}}\), so \(I_{rms}=\frac{V_{max}}{\sqrt{2}R}\)

Step2: Analyze the effect of changing \(L\) on \(I_{rms}\)

Since \(I_{rms}\) at resonance depends only on \(V_{max}\) (which is constant as per the problem) and \(R\) (which is given as a constant), changing \(L\) (while \(R\) and \(V_{max}\) are constant) has no effect on \(I_{rms}\)

Answer:

\(I_{rms}\) is unchanged.