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Step-by-step Solution
Step 1: List the given quantities
• Inductance, L = 30\text{ mH} = 30 \times 10^{-3}\text{ H}
• Resistance, R = 1\,\Omega
• Angular frequency, \omega = 300\,\text{rad/s}
• Phase angle by which current leads the voltage, \phi = 45^\circ
Step 2: Recall the phase angle formula for an LCR circuit
In a series LCR circuit, the phase angle \phi is given by
\tan \phi = \frac{X_C - X_L}{R},
where
X_C = \frac{1}{\omega C}\text{ (capacitive reactance)}, \quad X_L = \omega L\text{ (inductive reactance)}.
Step 3: Apply the given phase angle condition
Since \phi = 45^\circ , we know \tan 45^\circ = 1 . Hence,
1 = \frac{X_C - X_L}{R}.
This implies
X_C - X_L = R.
Step 4: Substitute the known values
Substitute X_C = \frac{1}{\omega C} , X_L = \omega L , and R = 1 :
\frac{1}{\omega C} - \omega L = 1.
Given \omega = 300 and L = 30 \times 10^{-3} , we have
\frac{1}{\omega C} - 300 \times 30 \times 10^{-3} = 1.
Calculate \omega L :
\omega L = 300 \times 30 \times 10^{-3} = 300 \times 0.03 = 9.
Thus,
\frac{1}{\omega C} - 9 = 1.
Therefore,
\frac{1}{\omega C} = 10.
Step 5: Solve for the capacitance C
From \frac{1}{\omega C} = 10 , we get
\omega C = \frac{1}{10} \;\; \Longrightarrow \;\; C = \frac{1}{10 \,\omega}.
Substituting \omega = 300 :
C = \frac{1}{10 \times 300} = \frac{1}{3000} = \frac{1}{3} \times 10^{-3}\,\text{F}.
Step 6: Identify the value of x
According to the question, the capacitance is given as
\frac{1}{x} \times 10^{-3}\,\text{F}.
We found C = \frac{1}{3} \times 10^{-3}\,\text{F} , so comparing these expressions shows x = 3 .
Final Answer
\boxed{3}
