6) A 1000 ? resistor, an inductor and a 0.30 F capacitor...

6) A 1000 ? resistor, an inductor and a 0.30 F capacitor are connected in series in a circuit and the resonant frequency is found to be 7000 Hz. What is the value of the inductor? A) 1.3 Hz D) 65 V G) 0.15 mH B) 7.2 mF E) 1.0 mH H) 0.31 F C) 1.7 mH F) 0.88 mH I) None of these 7) A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 T. What minimum current in the rod is needed in order for the magnetic force to cancel the weight of the rod? A) 1.2 A C) 9.8 A E) 4.9 A B) 2.5 A D) 3.7 A F) None of these 8) An ideal transformer has 40 turns on its primary coil and 360 turns on its secondary coil. If A voltage at 2.0 A is applied to the primary, by what factor does the voltage change in the secondary? A) Vs = 0.083 Vp D) Vs = 16 Vp G) Vs = 8.0 Vp B) Vs = 0.17 Vp E) Vs = 9.0 Vp H) None of these C) Vs = 12 Vp F) Vs = 15 Vp 9) An average of 120 kW of electric power is sent to a town from a power plant. The transmission lines have a total resistance of 0.40 ?. Calculate the power loss PLA if the power is transmitted at VA = 240 V, and the power loss PLB if the power is transmitted at VB = 24,000 V. A) PLA = 100 kW, PLB = 10 kW D) PLA = 100 kW, PLB = 10 W B) PLA = 10 kW, PLB = 10 W E) PLA = 100 W, PLB = 100 kW C) PLA = 10 kW, PLB = 100 kW F) None of these

Transcript

00:01 For this problem, you're considering a lot of circuits, transformers, things like that.

00:07 So there's going to be a lot of formulas that we use.

00:11 I'm going to write down all of the formulas as they come up.

00:16 In problem six, we are thinking about a series rlc circuit with a resistance of 1 ,000 oms, a capacitance of 0 .3 microferds, and a resonance frequency of 7 ,000 hertz.

00:38 And the question is, what is the inductance of the inductor in this circuit? so for an rlc circuit, the resonance frequency is related to the capacitance and the and the capacitance and the inductance in the circuit is equal to 1 over 2 pi times the square root of lc.

01:07 So this is the formula we want to use to solve this problem.

01:11 We're going to solve this formula for the inductance by first multiplying both sides of the equation by the square root of lc and dividing both sides by the resonance frequency.

01:25 So we're left with the square root of lc is equal to 1 over 2 pi times the resonance frequency.

01:33 We're then going to square both sides so that lc is equal to 1 over 4 pi squared times the resonance frequency squared.

01:43 And then we're going to divide both sides by the conductance, not conductance, capacitance, i'm sorry, to get the.

01:54 The inductance is equal to 1 over 4 pi squared resonance frequency squared times capacitance.

02:03 And at this point we can plug in the values given in the problem.

02:07 So l is going to be equal to 1 over 4 times pi squared times 7 ,000 hertz squared times.

02:18 And then you want to make sure to convert from micro ferrets to ferrets by multiplying by 10 to the negative 6.

02:30 And once you actually plug that in and calculate, you would get an inductance of 1 .72 times 10 to the negative 3 henry's, which the closest answer to that is 1 .7 millie henry answer choice c.

02:52 For problem 7, and i'm going to erase as i go so that each problem is done independently.

03:01 But for problem 7, we are considering a copper rod, and we're told a little bit of information about this rod.

03:10 We're told that it has a length of one meter and a mass of 0 .05 kilograms.

03:20 We're also told that it's in a magnetic field with a magnitude of 0 .1 tesla.

03:28 And we're asked what minimum current in the rod is needed in order for the magnetic force to cancel the weight of the rod.

03:37 So what we want is the magnetic force to be equal in magnitude to the weight of the rod.

03:47 The magnetic force for a current in a magnetic field is always given as the current times the length of the wire or rod in this case, times the magnetic field, times sign of the angle theta.

04:06 And then that must be equal to the weight, which is going to be mg.

04:13 Now we want to know the minimum current, and the minimum current will occur when sine of theta is at its maximum value of 1.

04:24 So i'm going to assume sine theta is equal to 1.

04:27 Which means i, lb is equal to mg.

04:32 And we can solve for the current i.

04:34 It's just going to be mg divided by lb.

04:38 And we can plug in the values given in the problem.

04:44 I use 9 .81 meters per second squared for g, assuming we're at the surface of the earth...

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