Alternating Current

JEE Main level, single correct, medium

21 of 190

Question 21 of 190

The equation of current in a purely inductive circuit is 5sin⁡(49πt−30∘)5\sin \left(49\pi t-30^{\circ}\right). If the inductance is 30mH30mH then the equation for the voltage across the inductor, will be

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Answer: (D) 23.1sin⁡(49πt+60∘)23.1\sin \left(49\pi t+60^{\circ}\right)

In inductive circuit, the voltage leads the current by 90∘90^{\circ}. Therefore,
ϕ0V−ϕ0I=90∘⇒ϕ0V=60∘{\phi }_{0V}-{\phi }_{0I}=90^{\circ} \Rightarrow {\phi }_{0V}=60^{\circ}
From the given equation 5sin⁡(49πt−30∘)5\sin \left(49\pi t-30^{\circ}\right), angular frequency ω=49πrads−1\omega =49\pi rad{s}^{-1}
The amplitude of the voltage across the inductor will be,
V0=i0XL{V}_{0}={i}_{0}{X}_{L}
⇒V0=i0(ωL)\Rightarrow {V}_{0}={i}_{0}\left(\omega L\right)
⇒V0=(5)(49π)(30×10−3)\Rightarrow {V}_{0}=\left(5\right)\left(49\pi \right)\left(30\times {10}^{-3}\right)=23.1=23.1
⇒V=23.1sin⁡(49πt+60∘)\Rightarrow V=23.1\sin \left(49\pi t+60^{\circ}\right)

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