Answer:
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
