Question
A particular inductor has appreciable resistance. When the inductor is connected to a 12 - V battery, the current in the inductor is 3.0 A. When it is connected to an AC source with an rms output of 12 V and a frequency of 60. Hz, the current drops to 2.0 A. What are (a) the impedance at 60. Hz and (b) the inductance of the inductor?
Step 1
The impedance (Z) is calculated by dividing the rms voltage (V) by the rms current (I). In this case, V = 12 V and I = 2.0 A. So, we have: \[Z = \frac{V}{I} = \frac{12 \, \text{V}}{2.0 \, \text{A}} = 6.0 \, \Omega\] Show more…
Show all steps
Your feedback will help us improve your experience
Averell Hause and 82 other Physics 102 Electricity and Magnetism educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A variable inductor is connected to a voltage source whose frequency can vary. The rms current is $I_{\mathrm{i}} .$ If the inductance is increased by a factor of 3.0 and the frequency is reduced by a factor of $2.0,$ what will be the new rms current in the circuit? The resistance in the circuit is negligible.
$\mathrm{A} 20 \mathrm{mH}$ inductor is connected across an AC generator that produces a peak voltage of $10 \mathrm{V}$. What is the peak current through the inductor if the emf frequency is (a) $100 \mathrm{Hz} ?$ (b) $100 \mathrm{kHz} ?$
An RLC series circuit consists of a $50-\Omega$ resistor, a $200-\mu \mathrm{F}$ capacitor, and a $120-\mathrm{mH}$ inductor whose coil has a resistance of $20 \Omega$. The source for the circuit has an ms emf of $240 \mathrm{V}$ at a frequency of $60 \mathrm{Hz}$. Calculate the ms voltages across the (a) resistor, (b) capacitor, and (c) inductor.
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD