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.

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: = 2цF = 2 x 10-6 F

Capacitance of the second capacitor = 7.4цF  = 7.4 x 10-6 F

Voltage of battery: V = 9 V  

Set Up the Problem:  

Capacitance of a series combination:  

+............

Capacitance of a series combination is given by:

Then the charge stored in the series combination is:  

Energy stored in the series combination is:  

Solve the Problem:  

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:

Capacitance of a parallel combination is given by

Then the charge stored in the parallel combination is

Energy stored in the parallel combination is:  

Solve the Problem:  

(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.

Given;

A 9.00V battery

A 2.00 μF capacitor ()

A 7.40 μF capacitor ()

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

= +

Substituting for the values of and in the above equation gives;

=> = +

=>   =

=> 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 = x C x

Substitute the values of V and C in the equation;

E = x 1.57 x

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 = +

Substituting for the values of and 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 = x C x

Substitute the values of V and C in the equation;

E = x 9.4 x

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.

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