Respuesta :
Explanation:
It is given that,
Resistance of the resistor, [tex]R=100\ \Omega[/tex]
Capacitance of the capacitor, [tex]C=0.1\ \mu F=0.1\times 10^{-6}\ F[/tex]
Inductance, [tex]L=2\ mH=2\times 10^{-3}\ H[/tex]
rms value of voltage, [tex]V_{rms}=120\ V[/tex]
The Ac source is operating at, [tex]f=\dfrac{1000}{\pi}\ Hz[/tex]
The inductive reactance of the inductor is given by :
[tex]X_L=\omega L[/tex]
[tex]X_L=2\pi fL[/tex]
[tex]X_L=2\pi \times \dfrac{1000}{\pi}\times 2\times 10^{-3}[/tex]
[tex]X_L=4\ \Omega[/tex]
The rms value of current is :
[tex]I_{rms}=\dfrac{V_{rms}}{R}[/tex]
[tex]I_{rms}=\dfrac{120}{100}[/tex]
[tex]I_{rms}=1.2\ A[/tex]
The rms voltage across the inductor is given by :
[tex]V_L=I_{rms}\times X_L[/tex]
[tex]V_L=1.2\ A\times 4\ \Omega[/tex]
[tex]V_L=4.8\ V[/tex]
So, the voltage across the inductor is 4.8 V. Hence, this is the required solution.
