Question

In Fig. 24.23, each capacitor has C = 4.90 µF, and Vab = +34.0 V.

Figure 24.23

(a) Calculate the charge on each capacitor.

Q1 = ___ C	Q2 =___ C
Q3 =___ C	Q4 =___ C

(b) Calculate the potential difference across each capacitor.

V1 = ___ V	V2 = ___V
V3 = ___V	V4 = ___V

(c) Calculate the potential difference between points a and d.
____V

Answer 1

Concepts and reason

In the given problem can be solved by using the series and parallel combination of capacitors. Need to identify the combinations of capacitors such as the parallel and series combination. Find the effective capacitance of each of these combinations and then need to find the equivalent capacitance of the entire circuit. Use the relation among the charge, capacitance, and voltage in order to find the charge on each capacitor. Use the relation among the charge, capacitance, and voltage in order to find the potential difference across each capacitor.

Fundamentals

The equivalent capacitance \(C_{s}\) of the capacitors connected in series can be calculated by using the following expression. \(\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}\)

Here, \(C_{1}, C_{2}\), and \(C_{3}\) are the capacitances of the capacitors. The equivalent capacitance \(C_{p}\) of the capacitors connected in parallel can be calculated by using the following expression. \(C_{p}=C_{1}+C_{2}+C_{3}\)

The relation among the charge \(Q\), capacitance \(C\), and the voltage \(V\) is expressed as follows:

\(Q=C V\)

\((\mathrm{a.1})\)

From the figure provided in the question, the capacitors \(C_{1}\) and \(C_{2}\) are in series. So, the effective capacitance of this combination of capacitors can be expressed as follows:

$$ \begin{array}{c} \frac{1}{C_{12}}=\frac{1}{C_{1}}+\frac{1}{C_{2}} \\ C_{12}=\frac{C_{1} C_{2}}{C_{1}+C_{2}} \end{array} $$

Given that the capacitance of each capacitor is same and is equal to \(\mathrm{C}\). Substitute \(4.90 \mu \mathrm{F}\) for both \(C_{1}\) and \(C_{2}\).

$$ \begin{aligned} C_{12}=& \frac{(4.90 \mu \mathrm{F})(4.90 \mu \mathrm{F})}{(4.90 \mu \mathrm{F})+(4.90 \mu \mathrm{F})} \\ &=2.45 \mu \mathrm{F} \end{aligned} $$

This combination of capacitors \(C_{1}\) and \(C_{2}\) is further connected to \(C_{3}\) in parallel. So, the equivalent capacitance of the combination of capacitors \(C_{1}, C_{2}\) and \(C_{3}\) can be expressed as follows:

\(C_{123}=C_{12}+C_{3}\)

Substitute \(2.45 \mu \mathrm{F}\) for \(C_{12}\) and \(4.90 \mu \mathrm{F}\) for \(C_{3}\).

$$ \begin{array}{c} C_{123}=(2.45 \mu \mathrm{F})+(4.90 \mu \mathrm{F}) \\ =7.35 \mu \mathrm{F} \end{array} $$

This combination of capacitors \(C_{1}, C_{2}\) and \(C_{3}\) is still connected to \(C_{4}\) in series. Thus, the equivalent capacitance of the given circuit is,

$$ \begin{array}{c} \frac{1}{C_{1234}}=\frac{1}{C_{123}}+\frac{1}{C_{4}} \\ C_{1234}=\frac{C_{123} C_{4}}{C_{123}+C_{4}} \end{array} $$

Substitute \(7.35 \mu \mathrm{F}\) for \(C_{123}\) and \(4.90 \mu \mathrm{F}\) for \(C_{4}\).

$$ \begin{aligned} C_{1234} &=\frac{(7.35 \mu \mathrm{F})(4.90 \mu \mathrm{F})}{(7.35 \mu \mathrm{F})+(4.90 \mu \mathrm{F})} \\ &=2.94 \mu \mathrm{F} \end{aligned} $$

The capacitors \(C_{1}\) and \(C_{2}\) are connected in series, and this series combination is connected to \(C_{3}\) in parallel and finally, this combination of capacitors is connected to \(C_{4}\) in series. Thus, the equivalent capacitance of the given circuit is calculated.

The relation among the charge \(Q\), capacitance \(C\), and the voltage \(V\) is expressed as follows:

\(Q=C V\)

The charge in total circuit is, \(Q_{\text {total }}=C_{1234} V_{a b}\)

Here, \(V_{a b}\) is the voltage across the circuit. Substitute \(2.94 \mu \mathrm{F}\) for \(C_{1234}\) and \(34.0 \mathrm{~V}\) for \(V_{a b}\).

\(\begin{aligned} Q_{\text {total }} &=(2.94 \mu \mathrm{F})(34.0 \mathrm{~V}) \\ &=99.96 \mu \mathrm{C} \end{aligned}\)

Thus, the charge in the total circuit is approximately \(100 \mu \mathrm{C}\).

The relation among the charge, capacitance, and voltage is used to find the total charge in the given circuit.

The combination of capacitors \(C_{1}, C_{2}\) and \(C_{3}\) is in series with the capacitance \(C_{4}\) and this entire combination is connected in series to the power source e (battery). Hence, the charge on the combined circuit of capacitors \(C_{1}, C_{2}\) and \(C_{3}\) is equal to the charge on the capacitor \(C_{4}\) and eventually equal to the total charge in the circuit. Thus, the charge on the capacitor \(C_{4}\) is \(100 \mu \mathrm{C}\). Therefore, \(Q_{4}=100 \mu \mathrm{C}\)

Thus, the charge on the capacitor \(C_{123}\) is \(100 \mu \mathrm{C}\). Therefore, \(Q_{123}=100 \mu \mathrm{C}\)

Now, use the relation among the charge, capacitance and voltage to find the voltage \(V_{123}\) across the capacitance \(C_{123}\). \(Q_{123}=C_{123} V_{123}\)

Solve the above equation for \(V_{123}\). \(V_{123}=\frac{Q 123}{C_{123}}\)

Substitute \(100 \mu \mathrm{C}\) for \(Q_{123}\) and \(7.35 \mu \mathrm{F}\) for \(C_{123}\).

\(V_{123}=\frac{100 \mu \mathrm{C}}{7.35 \mu \mathrm{F}}\)

\(=13.6 \mathrm{~V}\)

In a series combination of capacitors, the charge is the same on each capacitor. The relation among the charge, capacitance, and voltage are used to find the voltage across the combination of capacitors \(C_{1}, C_{2}\) and \(C_{3}\).

The combination of capacitors \(C_{1}\) and \(C_{2}\) is in parallel to \(C_{3}\). So, the voltage across the combination of capacitors \(C_{1}\) and \(C_{2}\) is same as the voltage across the capacitor \(C_{3} .\) And, this voltage must be equal to the voltage \(V_{123}\) across the total combination \(C_{1}, C_{2}\) and \(C_{3}\). Thus, the voltage across the capacitor \(C_{3}\) is, \(V_{3}=V_{123}\)

Substitute \(13.6 \mathrm{~V}\) for \(V_{123}\). \(V_{3}=13.6 \mathrm{~V}\)

Thus, the voltage across the combination of capacitors \(C_{1}\) and \(C_{2}\) is, \(V_{12}=V_{123}\)

Substitute \(13.6 \mathrm{~V}\) for \(V_{123}\). \(V_{12}=13.6 \mathrm{~V}\)

Again, use the charge, capacitance, voltage relation to find the charge on the capacitor \(C_{3}\). Thus, the charge on the capacitor \(C_{3}\) is expressed as follows:

\(Q_{3}=C_{3} V_{3}\)

Substitute \(4.90 \mu \mathrm{F}\) for \(C_{3}\) and \(13.6 \mathrm{~V}\) for \(V_{3}\).

\(Q_{3}=(4.90 \mu \mathrm{F})(13.6 \mathrm{~V})\)

\(=66.64 \mu \mathrm{C}\)

Thus, the charge on the capacitor \(C_{3}\) is approximately \(66.6 \mu \mathrm{C}\).

Taken the concern of that in parallel combination, the voltage across each capacitor is the same. And, the relation among the charge, voltage, and capacitance are used to find the charge across the capacitor \(C_{3}\).

From the given figure, the capacitors \(C_{1}\) and \(C_{2}\) are in series. So, the charge across each of these capacitors must be the same and is equal to the charge in the total combination of capacitors \(C_{1}\) and \(C_{2}\). The charge on the total combination of capacitors \(C_{1}\) and \(C_{2}\) is expressed as follows:

\(Q_{12}=C_{12} V_{12}\)

Substitute \(13.6 \mathrm{~V}\) for \(V_{12}\) and \(2.45 \mu \mathrm{F}\) for \(C_{12}\).

\(\begin{aligned} Q_{12}=&(2.45 \mu \mathrm{F})(13.6 \mathrm{~V}) \\ &=33.3 \mu \mathrm{C} \end{aligned}\)

As the capacitors \(C_{1}\) and \(C_{2}\) are in series, the charge on both these capacitors must be same and is equal to \(33.3 \mu \mathrm{C}\). Thus, \(Q_{1}=33.3 \mu \mathrm{C}\)

Taken the concern of that in series combination, the charge on each capacitor is same. And, the relation among the charge, voltage, and capacitance is used to find the charge on the capacitors \(C_{1}\) and \(C_{2}\)

\(\mathbf{a . 2}\)

As the capacitors \(C_{1}\) and \(C_{2}\) are in series, the charge on both these capacitors must be same and is equal to charge on the combination of the capacitors \(33.3 \mu \mathrm{C}\). Thus, \(Q_{2}=33.3 \mu \mathrm{C}\)

The charge on the capacitors is same when they are connected in series combination.

\(\mathbf{a . 3}\)

Thus, the charge on the capacitor \(C_{3}\) is expressed as follows:

\(Q_{3}=C_{3} V_{3}\)

Substitute \(4.90 \mu \mathrm{F}\) for \(C_{3}\) and \(13.6 \mathrm{~V}\) for \(V_{3}\).

$$ \begin{aligned} Q_{3}=&(4.90 \mu \mathrm{F})(13.6 \mathrm{~V}) \\ &=66.64 \mu \mathrm{C} \end{aligned} $$

Thus, the charge on the capacitor \(C_{3}\) is approximately \(66.6 \mu \mathrm{C}\).

The capacitor \(C_{3}\) and the combination of the capacitors \(C_{1}, C_{2}\) are in connected in parallel combination. Hence, the voltage across the capacitor \(C_{3}\) is equal to the voltage across the combination of the capacitors \(C_{1}, C_{2}\) and \(C_{3}\).

The charge on the capacitor \(C_{4}\) is equal to the charge on the resultant capacitance of the four capacitors. Therefore. \(Q_{4}=100 \mu \mathrm{C}\)

The combination of capacitors \(C_{1}, C_{2}\) and \(C_{3}\) is in series with the capacitance \(C_{4}\) and this entire combination is connected in series to the power source (battery). Hence, the charge on the combined circuit of capacitors \(C_{1}, C_{2}\) and \(C_{3}\) is equal to the charge on the capacitor \(C_{4}\) and eventually equal to the total charge in the circuit.

(b.1) The charge on the capacitor \(C_{1}\) is expressed as follows:

\(Q_{1}=C_{1} V_{1}\)

Rearrange the above relation for the voltage \(V_{1}\). \(V_{1}=\frac{Q_{1}}{C_{1}}\)

Substitute \(33.3 \mu \mathrm{C}\) for \(Q_{1}\) and \(4.90 \mu \mathrm{F}\) for \(C_{1}\).

$$ \begin{array}{c} V_{1}=\frac{33.3 \mu \mathrm{C}}{4.90 \mu \mathrm{F}} \\ =6.80 \mathrm{~V} \end{array} $$

The charge, capacitance, and voltage relation is used and calculated the values of voltages across the first capacitor.

\(\mathbf{b . 2}\)

The charge on the capacitor \(C_{2}\) is expressed as follows:

\(Q_{2}=C_{2} V_{2}\)

Rearrange the above relation for the voltage \(V_{2}\). \(V_{2}=\frac{Q_{2}}{C_{2}}\)

Substitute \(33.3 \mu \mathrm{C}\) for \(Q_{2}\) and \(4.90 \mu \mathrm{F}\) for \(C_{2}\).

$$ \begin{array}{c} V_{2}=\frac{33.3 \mu \mathrm{C}}{4.90 \mu \mathrm{F}} \\ =6.80 \mathrm{~V} \end{array} $$

The charge, capacitance, and voltage relation is used and calculated the values of voltages across the first capacitor.

\(\mathbf{b . 3}\)

The charge on the capacitor \(C_{3}\) is expressed as follows:

\(Q_{3}=C_{3} V_{3}\)

Rearrange the above relation for the voltage \(V_{3}\). \(V_{3}=\frac{Q_{3}}{C_{3}}\)

Substitute \(66.6 \mu \mathrm{C}\) for \(Q_{3}\) and \(4.90 \mu \mathrm{F}\) for \(C_{3}\).

\(V_{3}=\frac{66.6 \mu \mathrm{C}}{4.90 \mu \mathrm{F}}\)

\(=13.6 \mathrm{~V}\)

The charge, capacitance, and voltage relation is used and calculated the values of voltages across the first capacitor.

b. 4

The charge on the capacitor \(C_{4}\) is expressed as follows:

\(Q_{4}=C_{4} V_{4}\)

Rearrange the above relation for the voltage \(V_{4}\). \(V_{4}=\frac{Q_{4}}{C_{4}}\)

Substitute \(100 \mu \mathrm{C}\) for \(Q_{4}\) and \(4.90 \mu \mathrm{F}\) for \(C_{4}\).

\(V_{4}=\frac{100 \mu \mathrm{C}}{4.90 \mu \mathrm{F}}\)

\(=20.4 \mathrm{~V}\)

The charge, capacitance, and voltage relation is used and calculated the values of voltages across the first capacitor.

(c) The capacitors that are connected between the points \(a\) and \(\mathrm{d}\) are \(C_{1}, C_{2}\) and \(C_{3}\). The equivalent capacitance of the combination of capacitors \(C_{1}, C_{2}\) and \(C_{3}\) is \(C_{123}\) and is equal to \(7.35 \mu \mathrm{F}\). The combination of capacitors \(C_{1}\) and \(C_{2}\) is in parallel to \(C_{3} .\) So, the voltage across the combination of capacitors \(C_{1}\) and \(C_{2}\) is same as the voltage across the capacitor \(C_{3}\). And, this voltage must be equal to the voltage \(V_{123}\) across the total combination \(C_{1}, C_{2}\) and \(C_{3}\). Thus, the voltage across the capacitor \(C_{123}\) is equal to the voltage across \(C_{3}\). \(V_{123}=V_{3}\)

The voltage cross \(C_{3}\) is already calculated in part (b) and is equal to \(13.6 \mathrm{~V}\). Thus. \(V_{123}=13.6 \mathrm{~V}\)

As this capacitor \(C_{1}, C_{2}\) and \(C_{3}\) combination is connected between the points a and \(\mathrm{d}\), the voltage across this combination must be same as the voltage between the points a and \(\mathrm{d}\). Thus, \(V_{123}=V_{a d}\)

Therefore, the voltage between the points a and \(\mathrm{d}\) is, \(V_{a d}=13.6 \mathrm{~V}\)

The combination of capacitors \(C_{1}, C_{2}\) and \(C_{3}\) is connected in between the points a and d. So, the voltage across the total combination of capacitors \(C_{1}, C_{2}\) and \(C_{3}\) is equal to the voltage between the points a and \(\mathrm{d}\).

Part a.1

The charge on the first capacitor is \(33.3 \mu \mathrm{C}\).

Part a.2 The charge on the second capacitor is \(33.3 \mu \mathrm{C}\).

Part a.3 The charge on the third capacitor is \(66.6 \mu \mathrm{C}\).

Part a.4 The charge on the fourth capacitor is \(100 \mu \mathrm{C}\).

Part b.1

The voltage across the first capacitors is \(6.80 \mathrm{~V}\).

Part b.2

The voltage across the second capacitors is \(6.80 \mathrm{~V}\).

Part b.3

The voltage across the third capacitors is \(13.6 \mathrm{~V}\)

Part b.4

The voltage across the fourth capacitors is \(20.4 \mathrm{~V}\).

Part c Thus, the voltage between the points a and \(\mathrm{d}\) is \(13.6 \mathrm{~V}\).