Your parallel capacitors are 15 μf and 20 μf. The series capacitors are 10 μf and 12 μf. This circuit is connected to a 14 v battery, also determine the potential energy and the voltage across each capacitor

a. i. 6.608 V ii. 5.507 V iii. 1.89 V iv. 1.89 V

b. i. 0.22 mJ ii. 0.182 mJ iii. 0.027 mJ iv. 0.036 mJ

Explanation:

a. The voltage across each capacitor

Since the 15 μf and 20 μf capacitors are in parallel, their total capacitance is C = 15 μf + 20 μf = 35 μf.

Also, since C is in series with the 10 μf and 12 μf which are in series, their total capacitance, C' is gotten from 1/C' = 1/10 μf + 1/12 μf + 1/35 μf  

1/C' = (12 + 42 + 35)/420 /μf

1/C' = 89/420 /μf

C' = 420/89 μf

C' = 4.72 μf

The total charge in the circuit' is thus Q = C'V where V = voltage = 14 V

So, Q = C'V =  4.72 μf × 14 V = 66.08 μC

Since the 10 μf and 12 μf are in series, Q is the charge flowing through them.

Since Q = CV and V = Q/C

i. The voltage across the 10 capacitor is

V = 66.08 μC/10 μF = 6.608 V

ii. The voltage across the 12 capacitor is

V = 66.08 μC/12 μF = 5.507 V

The voltage across the 15 μF and 20 μF capacitors.

Since the capacitors are in parallel, the voltage across them is the voltage across their combined capacitance, C

So, V = Q/C = 66.08 μC/35 μF = 1.89 V

iii. The voltage across the 15 μF capacitor is 1.89 V

iv. The voltage across the 20 μF capacitor is 1.89 V

b. The potential energy of each capacitor

i. The potential energy of the 10 μF capacitor

E = 1/2CV² where C = Capacitance = 10 μF = 10 × 10⁻⁶ F and V = voltage across capacitor = 6.608 V

E = 1/2CV²

E = 1/2 × 10 × 10⁻⁶ F(6.608 V)²

E = 5 × 10⁻⁶ F(43.666) V²

E = 218.33 × 10⁻⁶ J

E = 0.21833 × 10⁻³ J

E = 0.21833 mJ

E ≅ 0.22 mJ

ii. The potential energy of the 12 μF capacitor

E = 1/2CV² where C = Capacitance = 12 μF = 12 × 10⁻⁶ F and V = voltage across capacitor = 5.507 V

E = 1/2CV²

E = 1/2 × 12 × 10⁻⁶ F(5.507 V)²

E = 6 × 10⁻⁶ F(30.327) V²

E = 181.96 × 10⁻⁶ J

E = 0.18196 × 10⁻³ J

E = 0.18196 mJ

E ≅ 0.182 mJ

iii. The potential energy of the 15 μF capacitor

E = 1/2CV² where C = Capacitance = 15 μF = 15 × 10⁻⁶ F and V = voltage across capacitor = 1.89 V

E = 1/2CV²

E = 1/2 × 15 × 10⁻⁶ F(1.89 V)²

E = 7.5 × 10⁻⁶ F(3.5721) V²

E = 26.79 × 10⁻⁶ J

E = 0.02679 × 10⁻³ J

E = 0.02679 mJ

E ≅ 0.027 mJ

iv. The potential energy of the 15 μF capacitor

E = 1/2CV² where C = Capacitance = 20 μF = 15 × 10⁻⁶ F and V = voltage across capacitor = 1.89 V

E = 1/2CV²

E = 1/2 × 20 × 10⁻⁶ F(1.89 V)²

E = 10 × 10⁻⁶ F(3.5721) V²

E = 35.721 × 10⁻⁶ J

E = 0.035721 × 10⁻³ J

E = 0.035721 mJ

E ≅ 0.036 mJ

note: the given question is incomplete.

complete question is given below.

which is an example of kinetic energy?

a. the energy stored in ethanol  

b. a hockey puck sliding across the ice  

c. a compressed spring  

d. a ball sitting at the top of a ramp

solution: the kinetic energy is the energy posses by the virtue of the motion of the object.

in the given options, a, c and d are the example of potential energy because it is the energy posses by the virtue of the rest of the object. but option b has the kinetic energy as the object is in motion.

hence, a hockey puck sliding across the ice   is the apex example of kinetic energy.

so, the option (b) a hockey puck sliding across the ice   is the correct option.