$A$ pure inductor of $25.0 \; mH$ is connected to a source of $220 \; V$. Find the inductive reactance and rms current in the circuit if the frequency of the source is $50 \; Hz$.
(A) The inductive reactance $X_L$ is given by the formula $X_L = 2 \pi \nu L$.
Given: $L = 25.0 \; mH = 25 \times 10^{-3} \; H$,$\nu = 50 \; Hz$,and $V_{rms} = 220 \; V$.
Substituting the values: $X_L = 2 \times 3.14 \times 50 \times 25 \times 10^{-3} \; \Omega = 7.85 \; \Omega$.
The rms current $I_{rms}$ is given by $I_{rms} = \frac{V_{rms}}{X_L}$.
Substituting the values: $I_{rms} = \frac{220 \; V}{7.85 \; \Omega} \approx 28.03 \; A$.
Given: $L = 25.0 \; mH = 25 \times 10^{-3} \; H$,$\nu = 50 \; Hz$,and $V_{rms} = 220 \; V$.
Substituting the values: $X_L = 2 \times 3.14 \times 50 \times 25 \times 10^{-3} \; \Omega = 7.85 \; \Omega$.
The rms current $I_{rms}$ is given by $I_{rms} = \frac{V_{rms}}{X_L}$.
Substituting the values: $I_{rms} = \frac{220 \; V}{7.85 \; \Omega} \approx 28.03 \; A$.
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