CUET Preparation Today
In a series L, C and R circuit with an ac source of frequency ω, the current leads the voltage by π/4. The value of capacitance C will be |
$ω(ω L-R)^{-1}$ $ω^{-1}(ω L-R)$ $(ω^2 L-R)^{-1}$ $(ω^2 L+ωR)^{-1}$ |
$(ω^2 L+ωR)^{-1}$ |
The correct answer is Option (4) → $(ω^2 L+ωR)^{-1}$ |
