Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
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
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
What are the gases that come to equilibrium between the upper layer of the ocean and the atmosphere?