Question

An electric lamp which runs at \[100\,{\text{V}}\] DC and consumes \[10\,{\text{A}}\] current is connected to AC mains at \[150\,{\text{V}}\], \[50\,{\text{Hz}}\] cycles with a choke coil in series. Calculate the inductance and drop of voltage across the choke. Neglect the resistance of the choke.

Answer 1

Hint: Use the relation between the voltage drop across resistor and the voltage drop across inductor. Use the relation between the reactance, frequency and current to determine the inductance of the choke.

Formula used:
The voltage drop \[V\] is given by
\[{V^2} = V_L^2 + V_R^2\] …… (1)
Here, \[{V_R}\] is the voltage drop across the resistor and \[{V_L}\] is the voltage drop across the inductor.
The voltage drop \[{V_L}\] across inductor is given by
\[{V_L} = I{X_L}\] …… (2)
Here, \[I\] is the current and \[{X_L}\] is the reactance.
The inductive reactance is given by
\[{X_L} = 2\pi fL\] …… (3)
Here, \[f\] is the angular frequency and \[L\] is the inductance.

Complete step by step answer:
An electric lamp which runs at \[100\,{\text{V}}\] DC and consumes \[10\,{\text{A}}\] current is connected to AC mains at \[150\,{\text{V}}\], \[50\,{\text{Hz}}\] cycles with a choke coil in series.
Calculate the drop of voltage \[{V_L}\] across the choke.
Rearrange equation (1) for the drop across inductor.
\[{V_L} = \sqrt {{V^2} - V_R^2} \]
Substitute \[150\,{\text{V}}\] for \[V\] and \[100\,{\text{V}}\] for \[{V_R}\] in the above equation.
\[{V_L} = \sqrt {{{\left( {150\,{\text{V}}} \right)}^2} - {{\left( {100\,{\text{V}}} \right)}^2}} \]
\[ \Rightarrow {V_L} = \sqrt {22500 - 10000} \]
\[ \Rightarrow {V_L} = \sqrt {12500} \]
\[ \Rightarrow {V_L} = 111.8\,{\text{V}}\]
Hence, the voltage drop across the choke is \[111.8\,{\text{V}}\].
Determine the inductance of the choke.
Substitute \[2\pi fL\] for \[{X_L}\] in equation (2).
\[{V_L} = I2\pi fL\]
\[ \Rightarrow {V_L} = 2\pi IfL\]
Rearrange the above equation for the inductance \[L\].
\[L = \dfrac{{{V_L}}}{{2\pi If}}\]
Substitute \[111.8\,{\text{V}}\] for \[{V_L}\], \[3.14\] for \[\pi \], \[10\,{\text{A}}\] for \[I\] and \[50\,{\text{Hz}}\] for \[f\] in the above equation.
\[L = \dfrac{{111.8\,{\text{V}}}}{{2\left( {3.14} \right)\left( {10\,{\text{A}}} \right)\left( {50\,{\text{Hz}}} \right)}}\]
\[ \Rightarrow L = 0.0356\,{\text{H}}\]
\[ \Rightarrow L = 0.036\,{\text{H}}\]
Hence, the inductance is \[0.036\,{\text{H}}\].
Hence, the inductance and drop of voltage across the choke are \[0.036\,{\text{H}}\] and \[111.8\,{\text{V}}\] respectively.

Note:
If the resistance of the choke is not neglected, one may get the results other than the obtained.
Also note that you have to remember all the formulas and how to apply the formula in the specific numericals.