Question

A choke coil is needed to operate an arc lamp at $ 160V(rms) $ and $ 50Hz $ . The arc lamp has an effective resistance of $ 5\Omega $ when running at $ 10A(rms) $ . Calculate the inductance of the choke coil. If the same arc lamp is to be operated on $ 160V $ DC, what additional resistance would be required. Compare the power losses in both the cases.

Answer 1

Hint : In order to solve this question, we are going to first find the value of the inductance from the RMS values of the current and the voltage and the resistance given. After that, the additional resistance is calculated , and also the power consumed by the lamp and therefore, the power loss.
For the coil of the resistance, $ R $ and the inductance $ L $ and the angular frequency $ \omega $ ,The root mean square current is given by the formula.
 $ {I_{rms}} = \dfrac{{{V_{rms}}}}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }} $
AC power consumed by the arc lamp
 $ {P_{ac}} = {V_{rms}}{I_{rms}}\cos \phi $

Complete Step By Step Answer:
For the coil of the resistance, $ R $ and the inductance $ L $ and the angular frequency $ \omega $ ,The root mean square current is given by the formula.
 $ {I_{rms}} = \dfrac{{{V_{rms}}}}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }} $
As it is given that, $ {V_{rms}} = 160V $
Frequency, $ f = 50Hz $
Resistance, $ R = 5\Omega $
Putting these values,
 $ 10 = \dfrac{{160}}{{\sqrt {{5^2} + {{\left( {2\pi \times 50} \right)}^2}{L^2}} }} $
Squaring both sides, we get,
 $ \Rightarrow 25 + {\left( {100\pi L} \right)^2} = {16^2} \\
   \Rightarrow 25 + {10^5}{L^2} = 256 \\
   \Rightarrow L = \sqrt {\dfrac{{231}}{{{{10}^5}}}} = 0.05H \\ $
Let $ {R_A} $ be the additional resistance required for operating the arc lamp with $ 160V $ DC source, then,
 $ 10 = \sqrt {\dfrac{{160}}{{{R_A} + 5}}} \\
   \Rightarrow {R_A} + 5 = 16 \\
  \therefore {R_A} = 16 - 5 = 11\Omega \\ $
AC power consumed by the arc lamp
 $ {P_{ac}} = {V_{rms}}{I_{rms}}\cos \phi $
Where,
 $ \cos \phi = \dfrac{R}{Z} = \dfrac{R}{{\sqrt {{R^2} + {\omega ^2}{L^2}} }} $
Putting these values, we get
 $ \cos \phi = \dfrac{5}{{16}} $
Thus, the power is calculated as
 $ {P_{ac}} = 160 \times 10 = 1600W $
 Thus, the power loss of the coil is:
 $ 1600 - 500 = 1100W $
Which is equal to $ 68.75\% $ efficiency.

Note :
Inductance is the property of an electric conductor or circuit that causes an electromotive force to be generated by a change in the current flowing which is calculated. And the power loss depends on the Root mean square voltage and current values and also the cosine of the angle $ \phi $ .