Skip to content
Question of 50

Q.An alternating e.m.f. E=E0sin⁡ωtE = E_0 \sin\omega t is applied across a pure inductor of inductance LL. Show mathematically that the current flowing through it lags behind the applied e.m.f. by a phase angle of π2\dfrac{\pi}{2}. What is its inductive reactance? OR An alternating e.m.f. E=E0sin⁡ωtE = E_0 \sin\omega t is applied across a pure capacitor of capacitance CC. Show mathematically that the current flowing through it leads behind the applied e.m.f. by a phase angle of π2\dfrac{\pi}{2}. What is its capacitive reactance?

Manipur CohsemCOHSEM Manipur Higher Secondary Board 2016Subjective· 5mImportance★★★★★
0% · 0/50 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Applying E=L dI/dtE=L\,dI/dt and integrating shows the current in a pure inductor is a −cos⁡ωt-\cos\omega t function, i.e. lagging the emf by 90∘90^\circ; XL=ωLX_L=\omega L.

Pure inductor circuit: Let the applied alternating emf be E=E0sin⁡ωtE = E_0\sin\omega t, connected across a pure inductor of inductance LL (zero resistance). By Kirchhoff's voltage law, the applied emf at every instant equals the self-induced back-emf of the inductor:

E=LdIdtE = L\dfrac{dI}{dt}

⇒dIdt=E0Lsin⁡ωt\Rightarrow \dfrac{dI}{dt} = \dfrac{E_0}{L}\sin\omega t

Integrating with respect to tt:

I=∫E0Lsin⁡ωt dt=−E0ωLcos⁡ωt  (+const.,=0 for a purely oscillatory current)I = \int \dfrac{E_0}{L}\sin\omega t\,dt = -\dfrac{E_0}{\omega L}\cos\omega t \;(+\text{const.,} = 0 \text{ for a purely oscillatory current})

Using −cos⁡ωt=sin⁡(ωt−π/2)-\cos\omega t = \sin(\omega t - \pi/2):

I=E0ωLsin⁡(ωt−π2)=I0sin⁡(ωt−π2),I0=E0ωLI = \dfrac{E_0}{\omega L}\sin\left(\omega t - \dfrac{\pi}{2}\right) = I_0\sin\left(\omega t - \dfrac{\pi}{2}\right), \qquad I_0 = \dfrac{E_0}{\omega L}

Comparing with E=E0sin⁡ωtE = E_0\sin\omega t, the current's phase is (ωt−π/2)(\omega t - \pi/2) while the emf's phase is ωt\omega t — so the current lags the emf by a phase angle of π/2\pi/2.

The inductive reactance (opposition offered by the inductor to a.c., analogous to resistance) is

XL=E0I0=ωLX_L = \dfrac{E_0}{I_0} = \omega L

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.