Respuesta :
Answer:
[tex]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[/tex]
[tex]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[/tex]
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: [tex]C_{1}[/tex]= 2цF = 2 x 10-6 F
Capacitance of the second capacitor [tex]C_{2}[/tex]= 7.4цF = 7.4 x 10-6 F
Voltage of battery: V = 9 V
Set Up the Problem:
Capacitance of a series combination:
[tex]\frac{1}{C_{s} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} }[/tex]+............
[tex]\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\\[/tex]
Capacitance of a series combination is given by:
[tex]C_{s}=\frac{Q}{V}[/tex]
Then the charge stored in the series combination is:
[tex]Q=C_{s} V[/tex]
Energy stored in the series combination is:
[tex]U_{c}=\frac{1}{2} V^{2} C_{s}[/tex]
Solve the Problem:
[tex]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[/tex]
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:
[tex]C_{p} =C_{1} +C_{2} +C_{3}[/tex]
[tex]C_{p} =2+7.4=9.4*10^-6F[/tex]
Capacitance of a parallel combination is given by
[tex]C_{p} =\frac{Q}{V}[/tex]
Then the charge stored in the parallel combination is
[tex]Q=C_{p} V[/tex]
Energy stored in the parallel combination is:
[tex]U_{c}=\frac{1}{2} V^2C_{p}[/tex]
Solve the Problem:
[tex]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[/tex]
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 ([tex]C_{1}[/tex])
A 7.40 μF capacitor ([tex]C_{2}[/tex])
(a) If the capacitors are connected in series, then different voltages pass across them and the total capacitance (C) is given by
[tex]\frac{1}{C}[/tex] = [tex]\frac{1}{C_{1} }[/tex] + [tex]\frac{1}{C_{2} }[/tex]
Substituting for the values of [tex]C_{1}[/tex] and [tex]C_{2}[/tex] in the above equation gives;
=> [tex]\frac{1}{C}[/tex] = [tex]\frac{1}{2}[/tex] + [tex]\frac{1}{7.4}[/tex]
=> [tex]\frac{1}{C}[/tex] = [tex]\frac{7.4 + 2}{7.4 * 2}[/tex]
=> 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 = [tex]\frac{1}{2}[/tex] x C x [tex]V^{2}[/tex]
Substitute the values of V and C in the equation;
E = [tex]\frac{1}{2}[/tex] x 1.57 x [tex]9^{2}[/tex]
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 = [tex]C_{1}[/tex] + [tex]C_{2}[/tex]
Substituting for the values of [tex]C_{1}[/tex] and [tex]C_{2}[/tex] 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 = [tex]\frac{1}{2}[/tex] x C x [tex]V^{2}[/tex]
Substitute the values of V and C in the equation;
E = [tex]\frac{1}{2}[/tex] x 9.4 x [tex]9^{2}[/tex]
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.
