If 20 amperes circulate in a closed circuit, how much charge passes any point in
ID: 1447945 • Letter: I
Question
If 20 amperes circulate in a closed circuit, how much charge passes any point in 5 sec? a. 0.25 C b. 4 C c. 20 C d. 100 C 2. A metal wire has a resistance of 1.0-Ohm. What will be the resistance of a wire made of the same material but twice as long and with half the cross-sectional area? a. 0.4-Ohm b. 2.0-Ohm c. 0.02-Ohm d. 4.0-Ohm 3. The rate at which energy is dissipated in a lamp that carries 0.5 A at 120 V is a. 1/6 watts b. 2 watts c. 60 watts d. 240 watts 4. A 12 ohm load is connected across a 6.0 V source. How much energy does it use in 1/2 an hour? I kWh =3.6 times 10^6 J a. 1.5 times 10^3 kWh b. 2.0 times 10^-3 kWh c. 3.0 times 10^-3 kWh d. 12.0 times 10^-3 kWh 5. Of the following, the copper conductor that has the least resistance is a. thin, long, and hot b. thick, short, and cool c. thick, long, and hot d. thin, short, and cool 6. If all the components of a circuit are connected in series, the physical quantity that is the same at all points is the a. voltage b. current c. resistance d. power 7. The ratio of the potential difference across a conductor and the current in the conductor is called the of the conductor. a. electrical potential b. conductance c. capacitance d. resistance 8. Two lamps, one with a thick filament and one with a thin filament of the same length and material, are connected in series to a battery. The voltage is greater across the lamp with the a. thick filament b. thin filament c. voltage is the same for both 9. Which statement is correct? a. Charge flows in a circuit. b. Voltage flows through a circuit. c. Resistance is established across a circuit. d. Current causes voltage.Explanation / Answer
here,
1)
current , I = 20 A
time , t = 5 s
charge , Q = I * t
Q = 20 * 5
Q = 100 C
the charge passed is d. 100 C
2)
resistance of wire , r0 = 1 ohm
l = 2 * l0 and area = area0 /2
new resistance , r = p * l / area
r = p * 2 * l0 /(area/2 )
r = 4 * p * l0/area0
r = 4 * r0 = 4 ohm
the new resistance is d. 4 ohm
3)
current , I = 0.5 A
voltage , V = 120 V
power , P = V * I
P = 0.5 * 120 = 60 W
the rate at which energy is dissapated is c. 60 W
4)
resistance , R = 12 ohm
voltage, V = 6 V
time , t = 0.5 h
energy , E = V^2 /R * t
E = 6^2 /12 * 0.5
E = 1.5 * 10^-3 KWh
the energy dissapated is a. 1.5 * 10^-3 KWh
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