The correct option is C B2 will be brighter than B1
Bulb acts as a resistor in circuit.
Therefore, IC=E√R2+(1ωC)2
From the given diagram, XC=1Cω=20π Ω
Brightness of B1=I2CR
And , IL=E√R2+(ωL)2
From the given diagram, XL=Lω=10×10−3×2π×50=πΩ
Thus, XC>XL⇒IL>IC
Therefore, B2 will be brighter.
Hence, option (b) is the correct answer.