If L = 25 mH, C=1mF and R = 10Ω  then R.M.S current across source is

XL=ωL=20025×10−3=5Ω​XC=1ωC=1200×10−3=5Ω∴XC=XL​​ Current across inductor and capacitor are out of phase and of same magnitude hence only impedance is same as resistance .⇒i=VZ=VR=5010=5A​Now,  irms=i2=52=3.54A