Students in science class designed the
electrical circuit shown in the figure below. The
circuit consists of a 12-volt (V) battery, a switch, a
capacitor (a two-plate electrical storage device), a
voltmeter (an instrument capable of measuring the
electrical potential difference between two points in
an electric circuit), and a resistor (a two-terminal
electronic device that resists the flow of electricity
in a circuit).
The students performed three experiments
using this electrical circuit.
Experiment 1
The students used a 10 megohm (MΩ)
resistor and a capacitor with a capacitance of 1
microfarad (µF). (Capacitance is the measure of the
maximum amount of electrical energy that a given
capacitor can store.) At the beginning of the ex-
periment (“time zero”) the capacitor was uncharged.
To begin the experiment, one student closed the
switch while another simultaneously started a timer.
At time zero, and at 10-second intervals thereafter, a
third student recorded the values from the
voltmeter. The results are set forth in Table 1.
Table 1
Time (sec) Voltmeter reading (V)
0
10
20
30
40 0.0
7.8
10.4
11.6
12.0
Experiment 2
The students used a 10 MΩ resistor and 5
different capacitors, each having a unique
capacitance. For each of the 5 capacitors, the
students used the voltmeter to determine the length
of time it took the uncharged capacitor to store 6
volts of electrical energy. The results are set forth
in Table 2.
Table 2
Capacitance (μF) Time for capacitor to
store 6 V of
electrical energy (sec)
1.2
1.0
0.6
0.4
0.2 7.2
6.0
3.6
2.4
1.2
Experiment 3
The students used a capacitor with a
capacitance of 1µF and 4 different resistors, each
having a unique resistance. For each of the 4
resistors, the students used the voltmeter to
determine the length of time it took the uncharged
capacitor to store 6 volts of electrical energy. The
results are set forth in Table 3.
Table 3
Resistance (MΩ) Time for capacitor to
store 6 V of
electrical energy (sec)
10.0
7.5
5.0
2.5 6.0
4.5
3.6
1.9
If the students had used a capacitor with a
capacitance of 2.0 µF in Experiment 2, the time
required for the capacitor to store 6 V of
electrical energy, based on the data in Table 2,
would have been closest to:
F.
8.0 sec.
G.
10.0 sec.
H.
12.0 sec.
J.
14.0 sec
To find the answer, we know that the time for a capacitor to store 6 V of electrical energy is directly proportional to the capacitance of the capacitor.
In Experiment 2, when the capacitance was 1.0 µF, the time taken was 6.0 sec.
If the capacitance had been 2.0 µF instead, the time taken can be calculated using the proportionality:
(new time) = (old time) * (new capacitance / old capacitance)
(new time) = 6.0 sec * (2.0 µF / 1.0 µF)
(new time) = 6.0 sec * 2.0
(new time) = 12.0 sec
Therefore, the time required for the capacitor to store 6 V of electrical energy with a capacitance of 2.0 µF would have been closest to 12.0 sec, which is option H.