Choke coil reduces AC current without wasting power, but a resistor does. Where does the energy go?

7 answers & replies

• [+][–]Nisha Verma12thSample· 1mo ago

  Nothing vanishes, it's borrowed and given back. In an inductor, the current rises in a quarter cycle and energy gets stored in the magnetic field ($\tfrac12 LI^2$). In the next quarter the field collapses and energy goes back to the source. Over one full cycle:

  $p = vi = V_0I_0\sin\omega t\cos\omega t = \tfrac12 V_0I_0\sin2\omega t$

  whose average over a cycle is zero. In a resistor the energy never comes back, it leaves as heat. So the choke "limits" the current by reactance and just shuffles energy.

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  • [+][–]Pooja Iyer12thSample· 1mo agoAuthor

    ahhh so the source gives energy to the field and the field gives it back, like a loan. 🥹 that finally makes sense. cos φ = 0 is "no net energy used", not "no current".

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    • [+][–]Nisha Verma12thSample· 1mo ago

      yep, and that's why we call the current there "wattless current". it flows but its average power is zero.

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• [+][–]Kavya NaikDropperSample· 1mo ago

  Another way to see why the choke is "free": in a series L–R circuit the voltage across the resistor and across the inductor are 90° out of phase, so they don't add up simply. $V^2 = V_R^2 + V_L^2$. A choke takes up most of the supply voltage as $V_L$, and that part of the voltage is in quadrature with the current, so it does no net work. A resistor in the same place would take up its share as $V_R$, which is in phase with the current, so it burns $I^2R$ all the time. Same drop in voltage, very different bills.

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• [+][–]Mehak AroraDropperSample· 1mo ago

  Your DC doubt, Pooja: an inductor opposes only changes in current. For steady DC, $\omega = 0$ so $X_L = \omega L = 0$ and the only opposition left is the small resistance of the wire. A choke on DC would basically be a short circuit and get very hot. That's why choke is only for AC (the current keeps changing so L keeps opposing).

  (Pooja, I had the same doubt in 11th. You're thinking in the right direction.)

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• [+][–]Farhan ShaikhDropperSample· 1mo ago

  Transformer losses, the four names:

  • Copper loss: resistance of the windings ($I^2R$). Thick wire for the high-current side.
  • Eddy currents in the core: reduced by laminating the core (thin insulated sheets).
  • Hysteresis: repeated magnetisation and demagnetisation. Use soft iron, which has a thin B–H loop.
  • Flux leakage: not all flux links the secondary. Wind coils on top of each other on a closed core.

  The humming of a transformer is another small loss (as sound). With good design, efficiency is 95%+.

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• [+][–]Tushar PawarDropperSample· 1mo ago

  Why a choke instead of a resistor in a real circuit, e.g. a tube light. With a resistor in series to cut the current, a lot of power would be wasted as heat, you pay for it on the bill. With a choke the average power is nearly zero (a real one has a small resistance so a very small loss), so it controls the current cheaply. Also the mnemonic for phase: ELI the ICE man. In an inductor (L), emf (E) leads current (I).

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