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Capacitors in circuits problems
1) You are given three capacitors of capacitances 6  F, 4  F and 3  F, respectively. List all the different
capacitances that can be generated by connecting these elements in various combinations.
For all the capacitors arranged in series (Figure 1A):
1
𝐶𝑇𝑜𝑡
1
1
1
= 6𝜇𝐹+4𝜇𝐹+3𝜇𝐹
=> 𝐶𝑇𝑜𝑡 = 1.3𝜇𝐹
For all of the capacitors arranged in parallel (Figure 1B):
𝐶𝑇𝑜𝑡 = 6𝜇𝐹 + 4𝜇𝐹 + 3𝜇𝐹 => 𝐶𝑇𝑜𝑡 = 13𝜇𝐹
For the combination of 1 in series with 2 in parallel (Figure 1C):
1
𝐶𝑇𝑜𝑡
1
𝐶𝑇𝑜𝑡
1
𝐶𝑇𝑜𝑡
=
1
1
+
=> 𝐶𝑇𝑜𝑡 = 2.13𝜇𝐹
3𝜇𝐹 4𝜇𝐹 + 6𝜇𝐹
=
1
1
+
=> 𝐶𝑇𝑜𝑡 = 2.77𝜇𝐹
4𝜇𝐹 6𝜇𝐹 + 3𝜇𝐹
=
1
1
+
=> 𝐶𝑇𝑜𝑡 = 3.23𝜇𝐹
6𝜇𝐹 3𝜇𝐹 + 4𝜇𝐹
For the combination of 2 in series in parallel with the last (Figure 1D):
𝐶𝑇𝑜𝑡 =
𝐶𝑇𝑜𝑡 =
𝐶𝑇𝑜𝑡 =
1
1
1
6𝜇𝐹 + 3𝜇𝐹
1
+ 6𝜇𝐹 = 7.7𝜇𝐹
1
1
+
3𝜇𝐹 4𝜇𝐹
+ 4𝜇𝐹 = 6.0𝜇𝐹
1
+ 3𝜇𝐹 = 5.4𝜇𝐹
1
1
+
4𝜇𝐹 6𝜇𝐹
2) A group of identical capacitors are connected first in series and then in parallel. The equivalent capacitance of
the parallel combination is 144 times larger than the equivalent capacitance of the series combination. How
many capacitors are there in the group?
1
1 𝐶𝑖
The total capacitance in series and in parallel can be as follows: 𝐶𝑠𝑒𝑟𝑖𝑒𝑠 = ∑𝑛
=
According to the problem 𝐶𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 = 144𝐶𝑠𝑒𝑟𝑖𝑒𝑠
𝐶
𝑛𝐶 = 144 𝑛 so 𝑛2 𝐶 = 144𝐶 which means there are 12 capacitors in this group.
𝐶
𝑛
𝑎𝑛𝑑 𝐶𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 = ∑𝑛1 𝐶𝑖 = 𝑛𝐶
3) Find the (a) equivalent capacitance and (b) the total charge and c) the total energy stored in the capacitance network
shown. Find also (d) the charge on each capacitor and (e) the voltage across each capacitor.
In order to solve this problem we need to simplify the circuit. The
following figure shows the different steps that you can take to find
the equivalent capacitance for the entire circuit:
a) 𝐶𝑒𝑞 = 8.6𝜇𝐹
b) 𝑄 = 𝐶𝑉 = (8.6𝜇𝐹)(12𝑉) = 103.4𝜇𝐶
1
1
c) 𝐸 = 2 𝑄𝑉 = 2 (103.4𝜇𝐶)(12𝑉) = 620𝜇𝐽
d) Let’s number the capacitors in the original circuit so that 1 and then 2 are going down on the left, while 3 then 4
are in the parallel path on the right, with 5 on the bottom. V1,2=V3,4,5=V=12V for either parallel path. Now we
can use the equation Q=CV to determine the charge in either side of the circuit.
a. 𝑄1,2 = 𝐶1,2 𝑉1,2 = (4𝜇𝐹)(12𝑉) = 48𝜇𝐶
b. 𝑄3,4,5 = 𝐶3,4,5 𝑉3,4,5 = (4.6𝜇𝐹)(12𝑉) = 55𝜇𝐶
e) Capacitors in series have the same charge, so we can use our answers in part d to answer part e.
a. 𝑄1 = 𝐶1 𝑉1 => 48𝜇𝐶 = 6𝜇𝐹𝑉1 => 𝑉1 = 8𝑉
b. 𝑄2 = 𝐶2 𝑉2 => 48𝜇𝐶 = 12𝜇𝐹𝑉2 => 𝑉2 = 4𝑉
c. 𝑄3 = 𝐶3 𝑉3 => 55𝜇𝐶 = 9𝜇𝐹𝑉3 => 𝑉3 = 6.13𝑉
d. 𝑄4 = 𝐶4 𝑉4 => 55𝜇𝐶 = 18𝜇𝐹𝑉4 => 𝑉4 = 3.06𝑉
e. 𝑄5 = 𝐶5 𝑉5 => 55𝜇𝐶 = 20𝜇𝐹𝑉5 => 𝑉5 = 2.76𝑉
You can check this by making certain that the voltage of the left two capacitors and of the right three capacitors
each add to 12 V total (which they do).
3) Consider a parallel-plate capacitor connected to a battery. How does the charge in this capacitor change if (a)
plate separation is doubled, (b) potential difference is doubled (by varying the voltage of the battery), and (c) if
both plate separation and potential difference are doubled? d) How does the energy stored change for a, b, and
c, above? Now, what if the same parallel plate capacitor is isolated from the battery? e) How does the charge
change if the plate separation is doubled? f) How does the energy stored change?
a. When connected to the battery: when d->2d, the capacitance decreases by a factor of 2 and since the
voltage can’t change, the charge must also decrease by a factor of 2.
b. If the potential difference is doubled, the capacitance remains unchanged (same capacitor), so the
charge must also double to keep C constant.
c. If the plate separation doubles and the potential doubles then the charge remains unchanged (part
a*part b or (0.5)*2=1)
1
1 𝑄2
𝐶
d. Energy storage 𝐸 = 2 𝐶𝑉 2 = 2
i. Half as large
ii. 4 times greater
iii. Twice as large
e. With an isolated capacitor, there is nowhere for the charge to go when the capacitance is changed, so
the charge is constant (the voltage across the plates would change)
f. If the capacitance is decreased by a factor of 2 with a constant charge, the energy must increase by a
factor of 2.