A resistor, an inductor and a capacitor are connected in series with a \[120\,{\text{V}}\], \[100\,{\text{Hz}}\] ac source. Voltage leads the current by \[35^\circ \] in the circuit. If the resistance of the resistor is \[10\,\Omega \] and the sum of the inductive and capacitive reactance is \[17\,\Omega \], calculate the self-inductance of the inductor.

Hint: Use the formula for phase difference between the voltage and current in the circuit. This formula gives the relation between the phase difference, inductive reactance, capacitive reactance and resistance. Calculate the value of inductive reactance using this equation and then use the formula for self-inductance in terms of inductive reactance and frequency.

Formulae used:
The phase difference \[\phi \] between the voltage and current is given by
\[\tan \phi = \dfrac{{{X_L} - {X_C}}}{R}\] …… (1)
Here, \[{X_L}\] is the inductive reactance, \[{X_C}\] is the capacitive reactance and \[R\] is the resistance.
The inductance \[L\] of an inductor is given by
\[L = \dfrac{{{X_L}}}{{2\pi f}}\] …… (2)
Here, \[{X_L}\] is the inductive reactance and \[f\] is the frequency.

Complete step by step solution:
We have given that the resistor, inductor and capacitor are connected in series in a circuit with potential difference \[120\,{\text{V}}\] and frequency \[100\,{\text{W}}\].
\[V = 120\,{\text{V}}\]
\[f = 100\,{\text{Hz}}\]
The phase difference between the voltage and current is \[35^\circ \].
\[\phi = 35^\circ \]
The resistance of the resistor is \[10\,\Omega \].
\[R = 10\,\Omega \]
The sum of the inductive and capacitive reactance is \[17\,\Omega \].
\[{X_L} + {X_C} = 17\,\Omega \] …… (3)
We have asked to calculate the self-inductance of the inductor. Let us first calculate the inductive reactance.Substitute \[35^\circ \] for \[\phi \] and \[10\,\Omega \] for \[R\] in equation (1).
\[\tan 35^\circ = \dfrac{{{X_L} - {X_C}}}{{10\,\Omega }}\]
\[ \Rightarrow {X_L} - {X_C} = 0.7 \times 10\]
\[ \Rightarrow {X_L} - {X_C} = 7\,{\text{Hz}}\] …… (4)
We can solve the equations (3) and (4) by adding them.
\[\left( {{X_L} + {X_C}} \right) + \left( {{X_L} - {X_C}} \right) = 17\,\Omega + 7\,{\text{Hz}}\]
\[ \Rightarrow 2{X_L} = 24\,\Omega \]
\[ \Rightarrow {X_L} = 12\,\Omega \]
Hence, the inductive reactance is \[12\,\Omega \].
Let is now calculate the self-inductance of the inductor.
Substitute \[12\,\Omega \] for \[{X_L}\], \[3.14\] for \[\pi \] and \[100\,{\text{Hz}}\] for \[f\] in equation (2).
\[L = \dfrac{{12\,\Omega }}{{2\left( {3.14} \right)\left( {100\,{\text{Hz}}} \right)}}\]
\[ \Rightarrow L = 0.019\,{\text{H}}\]
\[ \therefore L = 1.9 \times {10^{ - 2}}\,{\text{H}}\]

Hence, the self-inductance of the inductor is \[1.9 \times {10^{ - 2}}\,{\text{H}}\].

Note: There is no need to calculate the capacitive reactance in the circuit as we are asked to calculate the self-inductance of the inductor. The formula for the self-inductance includes only inductive reactance and not capacitive reactance. So, if one tries to calculate the capacitive reactance, it is a waste of time only.