a) The charge starts large and decreases to zero over time; the bulb starts brig
ID: 1446209 • Letter: A
Question
a) The charge starts large and decreases to zero over time; the bulb starts bright and dims over time.
b) The charge starts large and decreases to zero over time; the bulb starts dim and brightens, reaching a maximum brightness.
c) The charge and the brightness start at zero and continue to increase over time (never reaching a maximum value).
d)The charge starts at zero and increases to a maximum value; the bulb starts dim and brightens, reaching a maximum brightness.
e) The charge starts at zero and increases to some maximum value; the bulb starts bright and dims over time.
(b) Now open the bottom switch and close the top one. What happens to the charge on the capacitor and the brightness of the bulb as a function of time?
a) The charge decreases over time while the brightness of the bulb increases over time.
b)The charge and brightness both increase from zero over time.
c) The charge increases over time while the brightness of the bulb decreases over time.
d) The charge and brightness both decrease to zero over time.
(c) What statement below best characterizes when you will see the light bulb light (i.e., current in the circuit)?
a) The light bulb lights when the capacitor is charged.
b) The light bulb lights when the capacitor is uncharged.
c) The light bulb lights only when the capacitor is discharging.
d)The light bulb lights only when the capacitor begins charging or discharging.
e)The light bulb lights only when the capacitor is charging.
(d) From your observations, estimate the time scale for this circuit.
a) 5 seconds
b) 1/3 second
c) 10 seconds
d) 3 seconds
e)1 second
f) 1/5 second
(e) Now add the voltage chart (measure the voltage across C) and the current chart (measure the current through C). With the top switch open, close the bottom switch. What happens to the voltage across the capacitor & the current through the capacitor as a function of time?
a) The voltage across C increases from zero and saturates the the voltage of the battery; the current exponentially decays to zero.
b) The voltage across C increases from zero and saturates the the voltage of the battery; the current remains constant.
c) The voltage across C exponentially decays to zero; the current exponentially decays to zero.
d) The voltage across C exponentially decays to zero; the current starts at zero and saturates to a maximum value.
e) The voltage across C increases from zero and saturates the the voltage of the battery; the current starts at zero and saturates to a maximum value.
f) The voltage across C exponentially decays to zero; the current remains constant.
(f) Now open the bottom switch and close the top one. What happens to the voltage across the capacitor & the current through the capacitor as a function of time?
a) The voltage across C increases from zero and saturates the the voltage of the battery; the current exponentially decays to zero.
b) The voltage across C increases from zero and saturates the the voltage of the battery; the current starts at zero and saturates to a maximum value.
c) The voltage across C exponentially decays to zero; the current starts at zero and saturates to a maximum value.
d) The voltage across C exponentially decays to zero; the current remains constant.
e) The voltage across C exponentially decays to zero; the current exponentially decays to zero.
f) The voltage across C increases from zero and saturates the the voltage of the battery; the current remains constant.
(g) Control-click on the capacitor and change its C to 0.2 F. What happens to its time constant for charging/discharging?
a) The time constant is unchanged
b) The time constant halves
c) The time constant doubles
d) The time constant is reduced by a factor of 4
e) The time constant quadruples
Explanation / Answer
voltage, v = 25 V
Capacitnace of capacitor,c = 0.05 F
resistance of bulb, R = 10 ohms
(a) With the top switch open, close the bottom switch. What happens to the charge on the capacitor & the brightness of the bulb as a function of time?
When the capcitor is fully discharged = bulb brightness is high
as soon as capacitor is fully charged bulb brightness will be very low so, option e is correct,
The charge starts at zero and increases to some maximum value
the bulb starts bright and dims over time.
(b) Now open the bottom switch and close the top one. What happens to the charge on the capacitor and the brightness of the bulb as a function of time?
Same as Part a, ie
The charge increases over time while the brightness of the bulb decreases over time.
(c) What statement below best characterizes when you will see the light bulb light (i.e., current in the circuit)?
Since Brightnes decreases over time so current also dec over time therefore, option B is correct
(d) From your observations, estimate the time scale ? for this circuit.
Time scale is given as, t = R*C = 10 * 0.05 = 0.5 s or 1/2 s,
Please Check your options
(e) Now add the voltage chart (measure the voltage across C) and the current chart (measure the current through C). With the top switch open, close the bottom switch. What happens to the voltage across the capacitor & the current through the capacitor as a function of time?
Volatge decreases overtime while current inc over time, so option d is correct
(f) Now open the bottom switch and close the top one. What happens to the voltage across the capacitor & the current through the capacitor as a function of time?
The current and volatge both remain constant at zero value
so option e is correct.
(g) Control-click on the capacitor and change its C to 0.2 F. What happens to its time constant for charging/discharging?
time constant, t = R*C = 10 * 0.2
t = 2 s
option c is correct
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