Question 21.3: A Purely Inductive AC Circuit Goal Perform basic AC circuit ...
A Purely Inductive AC Circuit
Goal Perform basic AC circuit calculations for an inductive circuit.
Problem In a purely inductive AC circuit (see Active Fig. 21.6), L=25.0 \mathrm{mH} and the rms voltage is 1.50 \times 10^{2} \mathrm{~V}. Find the inductive reactance and rms current in the circuit if the frequency is 60.0 \mathrm{~Hz}.
Substitute L and f into Equation 21.8 to get the inductive reactance:
X_{L}=2 \pi f L=2 \pi\left(60.0 \mathrm{~s}^{-1}\right)\left(25.0 \times 10^{-3} \mathrm{H}\right)=9.42 \Omega
Solve Equation 21.9
\Delta V_{L,\mathrm{rms}}=I_{\mathrm{rms}}X_{L} (21.9)
for the rms current and substitute:
I_{\mathrm{rms}}=\frac{\Delta V_{L, \mathrm{rms}}}{X_{L}}=\frac{1.50 \times 10^{2} \mathrm{~V}}{9.42 \Omega}=15.9 \mathrm{~A}
Remark The analogy with DC circuits is even closer than in the capacitive case, because in the inductive equivalent of Ohm’s law, the voltage across an inductor is proportional to the inductance L, just as the voltage across a resistor is proportional to R in Ohm’s law.