Question

Suppose you have a 9.00 V battery, a 2.00 μF capacitor, and a 7.40 μF capacitor. (a) Find the charge and energy stored if the capacitors are connected to the battery in series. (b) Do the same for a parallel connection.

Answer 1

Answer:

$Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J$

$Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J$

Explanation:

a)

Identify the unknown:  

The charge and energy stored if the capacitors are connected in series  

List the Knowns:

Capacitance of the first capacitor: $C_{1}$= 2цF = 2 x 10-6 F

Capacitance of the second capacitor $C_{2}$= 7.4цF  = 7.4 x 10-6 F

Voltage of battery: V = 9 V  

Set Up the Problem:  

Capacitance of a series combination:  

$\frac{1}{C_{s} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} }$+............

$\frac{1}{C_{s} } =\frac{1}{2} +\frac{1}{ 7.4} \\C_{s} =\frac{2*7.4}{2+7.4}=1.575 *10^-6 F$

Capacitance of a series combination is given by:

$C_{s}=\frac{Q}{V}$

Then the charge stored in the series combination is:  

$Q=C_{s} V$

Energy stored in the series combination is:  

$U_{c}=\frac{1}{2} V^{2} C_{s}$

Solve the Problem:  

$Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J$

b)

Identify the unknown:  

The charge and energy stored if the capacitors are connected in parallel  

Set Up the Problem:  

Capacitance of a parallel combination:

$C_{p} =C_{1} +C_{2} +C_{3}$

$C_{p} =2+7.4=9.4*10^-6F$

Capacitance of a parallel combination is given by

$C_{p} =\frac{Q}{V}$

Then the charge stored in the parallel combination is

$Q=C_{p} V$

Energy stored in the parallel combination is:  

$U_{c}=\frac{1}{2} V^2C_{p}$

Solve the Problem:  

$Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J$

Answer 2

Answer:

(a) The charge and energy stored if the capacitors are connected to the battery in series are 14.13μC and 63.59μ J respectively.

(b) The charge and energy stored if the capacitors are connected to the battery in series are 84.6μC and 380.7μ J respectively.

Explanation:

Given;

A 9.00V battery

A 2.00 μF capacitor ($C_{1}$)

A 7.40 μF capacitor ($C_{2}$)

(a) If the capacitors are connected in series, then different voltages pass across them and the total capacitance (C) is given by

$\frac{1}{C}$ = $\frac{1}{C_{1} }$ + $\frac{1}{C_{2} }$

Substituting for the values of $C_{1}$ and $C_{2}$ in the above equation gives;

=> $\frac{1}{C}$ = $\frac{1}{2}$ + $\frac{1}{7.4}$

=>  $\frac{1}{C}$ = $\frac{7.4 + 2}{7.4 * 2}$

=> C = (7.4 x 2) / (7.4 + 2)

=> C = 1.57μF

(i) The charge (Q) stored is given by

Q = CV

Where;

V is the total voltage = 9.00V

C is the total capacitance = 1.57μF

Substituting for the values of V and C in the equation gives;

Q = 1.57μF x 9.00V

Q = 14.13μC

(ii) The energy (E) stored is given by

E = $\frac{1}{2}$ x C x $V^{2}$

Substitute the values of V and C in the equation;

E = $\frac{1}{2}$ x 1.57 x $9^{2}$

E = 63.59μ J

Therefore the charge and energy stored if the capacitors are connected to the battery in series are 14.13μC and 63.59μ J respectively.

(b) If the capacitors are connected in series, then same voltage passes across them and the total capacitance (C) is given by;

C = $C_{1}$ + $C_{2}$

Substituting for the values of $C_{1}$ and $C_{2}$ in the above equation gives;

=> C = 2 + 7.4

=> C = 9.4μF

(i) The charge (Q) stored is given by

Q = CV

Where;

V is the total voltage = 9.00V

C is the total capacitance = 9.4μF

Substituting for the values of V and C in the equation gives;

Q = 9.4μF x 9.00V

Q = 84.6μC

(ii) The energy (E) stored is given by

E = $\frac{1}{2}$ x C x $V^{2}$

Substitute the values of V and C in the equation;

E = $\frac{1}{2}$ x 9.4 x $9^{2}$

E = 380.7μ J

Therefore the charge and energy stored if the capacitors are connected to the battery in series are 84.6μC and 380.7μ J respectively.