An $LCR$ circuit has $L=10\ mH, R=3\Omega$, and $C =1\mu$ $F$ connected in series to a source of $15 \cos\omega t$ volt. The current amplitude at a frequency that is $10\%$ lower than the resonant frequency is:

Resonant frequency, $w_o=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{(10\times 10^{-3})(10^{-6})}}=10^4 rad/s$

New frequency, $\omega=(0.9)\omega_o=9\times 10^3 rad/s$

We have new ${ X }_{ L }=\omega L=9\times { 10 }^{ 3 }\times 10\times { 10 }^{ -3 }=90\Omega$ and $X_C= \dfrac {1}{\omega C}=111.11$ $ohm$.

Thus we calculate new $Z$ as

$\sqrt{3^2+[90-111.11]^2}$

$=21.32 \Omega$

Current amplitude $=\dfrac{V_o}{Z}=\dfrac{15}{21.32}=0.7 A$