Question

What capacitor in series with a 100 Ohm resistor and a 20.0 mH inductor will give a resonance frequency of 800 Hz? A series RLC circuit consists of a 99.0 Ohm resistor, a 0.130 H inductor, and a 20.0 muF capacitor. It is attached to a 120 V/60 Hz power line. What is the peak current I? The tuning circuit in an FM radio receiver is a series RLC circuit with a 0.2000 muH inductor. The receiver is tuned to a station at 105.6 MHz. What is the value of the capacitor in the tuning circuit? In the figure at right, a piece of wire is used to connect points a and b. How does that affect the brightness of the light bulbs? A resistor is connected to an ideal ac power supply. How does the average power dissipated in the resistor change as the frequency in the ac power supply decreases? Does it increase or decrease? Explain. An emf of 0.50 V is induced in a 1,500-turn coil at an instant when the current is 10 A and is changing at a rate of 130 A/s. Calculate the magnetic flux through each turn of the coil.

Answer 1

1. Resonance frequency for series RLC circuit

   $f = \frac{1}{2\pi \sqrt{LC}}$

$C = \frac{1}{4\pi ^{2}f^{2}L}=\frac{1}{4\pi ^{2}(800)^{2}(20\times 10^{-3})}=1.2F$

2.

Inductor impedence = $2\pi fL=2\pi (60)(0.130)=49.0$

capacitor impedence = $-j/2\pi fC=-j/(2\pi (60)(20.0\times 10^{-6}))=132.6$

Total impedence $Z = R+2\pi fL-j/2\pi fC= 99.0+49.0-132.6=15.4\Omega$

Peak current

$I = V/Z= 120V/15.4\Omega =7.79A$

3.

$C = \frac{1}{4\pi ^{2}f^{2}L}=\frac{1}{4\pi ^{2}(105.6)^{2}(0.2\times 10^{-6})}=11.35F$

4.

R1 becomes brighter. Connecting a wire from a to b provides a nearly zero resistance path from b to c and decreases the total resistance of the circuit from R1 + R2 to just R1. Ignoring internal resistance, the potential difference maintained by the battery is unchanged while the resistance of the circuit has decreased. The current passing through bulb R1 increases, causing this bulb to glow brighter. Bulb R2 goes out because essentially all of the current now passes through the wire connecting a and b and bypasses the filament of Bulb R2.

5.

Average power dissipated in the resistor circuit during one cycle

$P =\frac{E_{rms}^{2}}{R}$

It does not change with the frequency

6.

$\varepsilon =-L\frac{\Delta I}{\Delta t}=-(N\phi _{B}/I)\left ( \frac{\Delta I}{\Delta t} \right )$

$\phi =-\frac{0.5(10)}{(1500)(130)}=-2.56\times 10^{-5}Wb$