
A variable inductor is connected to an AC source. When inductance increases. Then reactance
A. And current remain same
B. Increases but current remains same
C. Increases but current decreases
D. Increases and current also increases
Answer
567.3k+ views
Hint: In order to solve this problem, assume an inductor of some inductance. For this, you can give the relation between the reactance of the inductor and inductance. This will give you an idea of how reactance is changing with respect to the inductance. Similarly, find the relation between inductor current and reactance then you can know how the current varies in the inductor.
Formula used:
$\eqalign{
& {X_L} = L\omega \cr
& {V_L} = L\dfrac{{d{i_L}}}{{dt}} \cr
& {I_m} = \dfrac{{{V_m}}}{{{X_L}}} \cr} $
Complete answer:
Let us consider that initially, the inductance of the inductor coil is L. They’ve given in the question that L has been increased. Let us say that the new inductance of the inductor coil is L’. They’ve given that, $L' > L$.
Now the reactance of an inductor coil is given by ${X_L} = L\omega $. Where L is the inductance and ω is the angular frequency. Now, reactance XL is directly proportional to L. When the ω is kept constant, the reactance XL will increase with an increase in L.
If the current in the circuit ${i_L}$ is given by
${i_L} = {I_m}\sin \left( {\omega t + \theta } \right)$
Where,
Im is the peak current
The voltage across the inductor will be given by
${V_L} = L\dfrac{{d{i_L}}}{{dt}} = L\dfrac{{d{I_m}\sin \left( {\omega t + \theta } \right)}}{{dt}} = {I_m}L\omega \cos \left( {\omega t + \theta } \right)$
So, the peak voltage is
$\eqalign{
& {V_m} = {I_m} \times L\omega \cr
& \Rightarrow {I_m} = \dfrac{{{V_m}}}{{L\omega }} \cr
& \Rightarrow {I_m} = \dfrac{{{V_m}}}{{{X_L}}} \cr} $
We have seen that the reactance of the inductor increases with the increase in inductance. Here, the current is inversely proportional to the reactance of the coil. This means that with an increase in impedance the current will decrease.
Therefore, with an increase in the inductance the reactance increases, and the current in the circuit decreases, when the voltage is kept constant.
So, the correct answer is “Option C”.
Note:
You can also solve this problem by assuming the values for the inductance and the ac voltage across the inductor. It will get you through the confusion and give you a clear idea of how they are changing. Here, you must also remember how the voltage and current of an inductor are related, to avoid ambiguity as the quantities are not directly mentioned in the problem.
Formula used:
$\eqalign{
& {X_L} = L\omega \cr
& {V_L} = L\dfrac{{d{i_L}}}{{dt}} \cr
& {I_m} = \dfrac{{{V_m}}}{{{X_L}}} \cr} $
Complete answer:
Let us consider that initially, the inductance of the inductor coil is L. They’ve given in the question that L has been increased. Let us say that the new inductance of the inductor coil is L’. They’ve given that, $L' > L$.
Now the reactance of an inductor coil is given by ${X_L} = L\omega $. Where L is the inductance and ω is the angular frequency. Now, reactance XL is directly proportional to L. When the ω is kept constant, the reactance XL will increase with an increase in L.
If the current in the circuit ${i_L}$ is given by
${i_L} = {I_m}\sin \left( {\omega t + \theta } \right)$
Where,
Im is the peak current
The voltage across the inductor will be given by
${V_L} = L\dfrac{{d{i_L}}}{{dt}} = L\dfrac{{d{I_m}\sin \left( {\omega t + \theta } \right)}}{{dt}} = {I_m}L\omega \cos \left( {\omega t + \theta } \right)$
So, the peak voltage is
$\eqalign{
& {V_m} = {I_m} \times L\omega \cr
& \Rightarrow {I_m} = \dfrac{{{V_m}}}{{L\omega }} \cr
& \Rightarrow {I_m} = \dfrac{{{V_m}}}{{{X_L}}} \cr} $
We have seen that the reactance of the inductor increases with the increase in inductance. Here, the current is inversely proportional to the reactance of the coil. This means that with an increase in impedance the current will decrease.
Therefore, with an increase in the inductance the reactance increases, and the current in the circuit decreases, when the voltage is kept constant.
So, the correct answer is “Option C”.
Note:
You can also solve this problem by assuming the values for the inductance and the ac voltage across the inductor. It will get you through the confusion and give you a clear idea of how they are changing. Here, you must also remember how the voltage and current of an inductor are related, to avoid ambiguity as the quantities are not directly mentioned in the problem.
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