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

reserve is the answer

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)