A defibrillator uses 120 mu F capacitor with a charge of 0.13C. What is the init

Question

A defibrillator uses 120 mu F capacitor with a charge of 0.13C. What is the initial voltage of the defibrillator? How much energy is stored in the defibrillator? If it discharges in a time of 4.2ms, what is the average current delivered to the torso? Three capacitors are connected as shown to a 24V potential voltage source. What is the total capacitance of this circuit? Cl = 3mF, C2 = C3 = 8mF Light bulbs A and B are identical in all ways except that B's filament is thicker than A's. If screwed into a 120V socket, A will be brighter because it has the most resistance B will be brightest because it has the most resistance A will be brighter because it has the least resistance B will be brightest because it has the least resistance Both will have the same brightness In the following circuit, how much power is dissipated by the 20 Q resistor? 1.67 W 5.00 W 7.50 W The filament in a light bulb is a resistor in the form of a thin piece of wire. The wire becomes hot enough to emit light because of the current in it. A bulb is connected to a 3.0 V battery and has 0.040A flowing through it, what is the resistance of the bulb? Which one of the following statements concerning resistance is true? The resistance of a semiconductor increases with temperature. Resistance is a property of resistors, but not conductors. The resistance of a metal wire changes with temperature. The resistance is the same for all samples of the same material. The resistance of a wire is inversely proportional to the length of the wire. 22. Determine the potential difference across the ends of a wire of resistance of 5.0 2 if 720C of charge passes through it per minute.

Explanation / Answer

16.

a)

Initial Voltage

V=Q/C =0.13/(120*10-6)

V=1083.33 V

b)

Energy stored

E=(1/2)CV2 =(1/2)*(120*10-6)*1083.332

E=70.4 J

c)

Average current

I=Q/t =0.13/(4.2*10-3)

I=30.95 A

17.

equivalent capacitance

Ceq=C1+C2*C3/(C2+C3)

Ceq=3+8*8/(8+8) =7 mF

so ANswer is B

18)

Answer is D ie B will be brightest beacuse it has least resistance

B is thicker means more diameter

A=pid2/4

so area is more

R=pL/A

=>Resistance is less for bulb B compared to bulb A

19.

In parallel voltage across both resistors is same ,so

V20=V40=10 V

so power dissipated in 20 ohms is

P=V2/R =102/20

P=5 W

so Answer is B

20)

Resistance of the bulb

R=Voltage/current =3/0.04

R=75 ohms

21)

Answer is C ,i.e The reistance of metal wire changes with temperature

22)

Given

charge Q=720 C

and time t=1 minute=60s

Current passing through the wire

I=Q/t=720/60 =12 A

Potential difference across wire

V=IR =12*5

V=60 V

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