what is the changlaced in series with a battery as shown, once the capucitors ar
ID: 1582847 • Letter: W
Question
what is the changlaced in series with a battery as shown, once the capucitors are folly charged, A) 7.2 x 10 c B) 1.3 x 10 c C) 2.3 x 10 c D) 1.6 x 10sc E) 3.2x 10 c V-12V Ci Ca 12) A uncharged capacitor is place in series with a resistor and an open switch that is connected to a battery (de voltage), when the switch is closed at t- 0 seconds, what is the charge on the capacitor after 2 seconds? Switch A) 8.0 x 10 C B) 1.8 x 10 c C) 50x 10 c D) 1.1 x 10 C E) 3.6x 10 C V- 20 V C= 4.0 F 13) A battery is connected in series with a switch, an un-charged capacitor and a light bulb, when switch is closed what happens to the brightness of the light bulb over time? Switch A) It keeps getting brighter then it stays the same brightness B) It starts out bright then gets dimmer until it goes out C) It starts out bright then gets dim until the capacitor is charged then it gets bright again D) It is very dim until the capacitor is charged then it is bright E) Once the switch is closed it turns on and never changes brightness Light bulb a 10 resistor, what is the internal resi the battery when the switch is closed and a 0.7 A current flows at the same time the voltmeter reads 6. 14) A 9 V battery (Vent-9V) is wired in series with across the battery? A)3.1 B) 1.9 C) 10 D) 7.3 E) 5.6 3 of 7Explanation / Answer
given
11.
C1 = 2.0*10^-6 F, C2 = 4*10^-6 F
V = 12 V
when capacitors are in series the net capacitance is 1/C = 1/C1+1/C2
C = c1*C2/(C1+C2)
C = (2*10^-6*4*10^-6)/(2*10^-6+4*10^-6) F
C = 1.33*10^-6 F
in series combination the charge is same
Q = C*V
Q = 1.33*10^-6*12 C
Q = 1.596*10^-5 C
Q = 1.6*10^-5 C
answer is option D
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12.
Charging of a capacitor in RC circuit
given
V = 20 V
C = 4.0*10^-6 F
R = 20000 ohm
t = 2 s
Q0 = C*V
Q0 = 4.0*10^-6*20 C = 8*10^-5 C
q(t) = Q0 (1-e^-t/R*C)
q(t) = 8*10^-5 ( 1-e^(-2/(20000*4*10^-6))) C
q(t) = 7.9999999998889*10^-5 C
q(t) = 8*10^-5 C
answer is option A
13. OPTION C is the appropriate one
when the switch is closed the potential difference will be large across capacitor and the gradually decreases later that would applear across the bulb
and the potentila will be divided across the capacitor and bulb initially later the total potential will be across the bulb
14. V = i(R+r)
i = V/(R+r)
0.7 = 9 /(10+r)
r = 2.86 ohm
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