Question

For the system of capacitors shown in the figure below(Figure 1), a potential difference of 25.0V is maintained across ab.

Part (a): What is the equivalent capacitance of this system between a and b in nF?

Part (b): How much charge is stored by this system in nC?

Part (c): How much charge does the 6.50 nF capacitor store in nC?

Part (d): What is the potential difference across the 7.50 nF capacitor in V?

Answer 1

Concept and Reason

The main concept used is the parallel and series combination of the capacitors and relation between charge, capacitance, and voltage.

The equivalent capacitance of the capacitance of the system is calculated by using the series and parallel combination of the capacitors. Charge stored in capacitors connected in series is same. The potential difference across the capacitors connected in parallel is same.

Fundamental

The capacitance of a system is given by,

$C = \frac{q}{V}$

Here, C is the capacitance of the capacitor, V is the potential difference across the capacitor and q is the charge stored in the capacitor.

If capacitors are connected in series, their total capacitance is given by inverse of sum of inverse of each capacitor connected, i.e.,

$\frac{1}{{{C_{{\rm{tot}}}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + ...... + \frac{1}{{{C_n}}}$ ,

Here, C_tot is the total capacitance and n is the number of capacitors connected. The charge across each capacitor is equal

If capacitors are connected in parallel, their total capacitance is given by sum if capacitance of each capacitor, i.e.,

${C_{{\rm{tot}}}} = {C_1} + {C_2} + ..... + {C_n}$ ,

Here, ${C_{{\rm{tot}}}}$ is the total capacitance and n is the number of capacitors connected. The voltage across each capacitor is equal.

(a)

The network of capacitors is,

The expression for finding total capacitance ( ${C_{{\rm{tot}}}}$ ) of capacitors connected in series is

$\frac{1}{{{C_{{\rm{tot}}}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + ...... + \frac{1}{{{C_n}}}$

Let the first capacitance 18.0 nF be C₁, the second capacitance 30.0nF be C₂, the third capacitance 10.0nF be C₃ and the equivalent capacitance be ${C_{{\rm{tot1}}}}$ , we get

$\begin{array}{c}\\\frac{1}{{{C_{{\rm{tot1}}}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\\\\ = \frac{1}{{18.0}} + \frac{1}{{30.0}} + \frac{1}{{10.0}}\\\\ = \frac{{17}}{{90}}\\\end{array}$

Take the reciprocal of it,

$\begin{array}{c}\\{C_{{\rm{tot1}}}} = \frac{{90}}{{17}}\\\\ = 5.29{\rm{ nF}}\\\end{array}$

The three capacitors in the middle row can be replaced by a single capacitor of $5.29{\rm{ nF}}$ .

Capacitance of all capacitors connected in parallel can be given by,

${C_{{\rm{tot}}}} = {C_1} + {C_2} + ..... + {C_n}$

Let the 1^st row capacitance 7.5 nF be C_4, 2nd row capacitance $5.29{\rm{ nF}}$ is ${C_{{\rm{tot1}}}}$ , 3rd row capacitance 6.5nF be C₅ and the total capacitance be ${C_{{\rm{tot}}}}$ , we get

$\begin{array}{c}\\{C_{{\rm{tot}}}} = {C_4} + {C_{{\rm{tot1}}}} + {C_5}\\\\ = 7.5 + 5.29 + 6.5\\\\ = 19.29{\rm{ nF}}\\\end{array}$

[Part a]

Part a

(b)

The charge q stored by a capacitor of capacitance C is given by

$q = CV$

Here, $V$ is the potential difference across the capacitor.

The total capacitance of this system is $19.29{\rm{ nF}}$ . It is given that potential difference across the system is $25{\rm{V}}$ .

The charge stored by capacitor is given by

$\begin{array}{c}\\q = CV\\\\ = \left( {19.29 \times {{10}^{ - 9}}{\rm{F}}} \right)\left( {25{\rm{V}}} \right)\\\\ = 482.25 \times {10^{ - 9}}{\rm{C}}\\\\ = 482.25{\rm{ nC}}\\\end{array}$

(c)

The charge q stored by a capacitor of capacitance C is given by

$q = CV$

Here, V is the potential difference across the capacitor.

The capacitance of the capacitor is 6.50 nF. The voltage across the capacitor is 25V. Charge stored in the capacitor will be given by,

$\begin{array}{c}\\q = CV\\\\ = 6.5{\rm{nF}} \times 25{\rm{V}}\\\\{\rm{ = 162}}{\rm{.5nC}}\\\end{array}$

[Part c]

Part c

(d)

As the 7.50 nF capacitor is in parallel to other capacitive systems, the potential difference across the capacitor will be same as the voltage across rest of the capacitive systems.

In this case, the potential difference across each capacitive system is 25 V.

The potential difference across the $7.50 {\rm{nF}}$ capacitor is also 25V.

Ans: Part a

The equivalent capacitance of this system is $19.29{\rm{ nF}}$ .