Question

The self inductance of a choke coil is 10 mH. When it is connected with a 10 V DC source then the loss of power is 20 watts. When it is connected with 10 volt AC source loss of power is 10 watts. The frequency of AC source will be:
(A) 50 Hz
(B) 60 Hz
(C) 80 Hz
(D) 100 Hz

Answer 1

Hint: To answer this question, we need to first write down the value of power for both DC and AC sources. From there we can derive the expression of resistance. We have to put the values from the question to the expression to get the required value of resistance. At the end, we need to put the values in the formula of impedance. This will give us the required answer, for this question.

Complete step by step answer:
We should know that with power, that is expressed in P, the formula is:
$P = \dfrac{{{V^2}}}{R}$
Hence R can be expressed as:
$R = \dfrac{{{V^2}}}{P}$
Now put the values of V and P from the question in the expression to get R:
$R = \dfrac{{{{10}^2}}}{{20}} = 5\Omega $
In case of AC power, the power is expressed as:
$P = \dfrac{{V_{rms}^2R}}{{{Z^2}}}$
From the expression Z is expressed as:
${Z^2} = \dfrac{{{{10}^2} \times 5}}{{10}} = 50\Omega $
We know that the impedance is given as:
${Z^2} = {R^2} + 4{\pi ^2}{v^2}{L^2}$
Put the values in the above expression to get that:
$
  50 = {5^2} + {3.14^2} \times {v^2} \times {(10 \times {10^{ - 3}})^2} \\
   \Rightarrow v = 80Hz \\
 $

So we can say that the frequency of AC source will be 80 Hz. Hence the correct answer is option C.

Note: In this question, we have come across the term of self inductance. By self inductance we mean the induction of a voltage in case of a current carrying wire, when the current in the wire changes by itself. In the situation of self inductance the magnetic field is created by a changing current in the circuit by itself so as to induce a voltage in the same circuit. This voltage is thus termed as self-induced.