hw16-solutions

# hw16-solutions - marout(ssm979 – hw16 – turner...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: marout (ssm979) – hw16 – turner – (56725) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The circuit has been connected as shown in the figure for a “long” time. 48 V S 13 μ F 7 Ω 9 Ω 1 Ω 4 7 Ω What is the magnitude of the electric po- tential across the capacitor? Correct answer: 20 V. Explanation: Let : R 1 = 7 Ω , R 2 = 9 Ω , R 3 = 1 Ω , R 4 = 47 Ω , and C = 13 μ F = 1 . 3 × 10 − 5 F . E S C t b a b I t R 1 I t R 2 I b R 3 I b R 4 “After a long time” implies that the capac- itor C is fully charged, so it acts as an open circuit with no current flowing to it. The equivalent circuit is I t R 1 I t R 2 R 3 I b I b R 4 a b R t = R 1 + R 2 = 7 Ω + 9 Ω = 16 Ω and R b = R 3 + R 4 = 1 Ω + 47 Ω = 48 Ω , so I t = E R t = 48 V 16 Ω = 3 A and I b = E R b = 48 V 48 Ω = 1 A . Across R 1 , E 1 = I t R 1 = (3 A) (7 Ω) = 21 V and across R 3 E 3 = I b R 3 = (1 A) (1 Ω) = 1 V . Since E 1 and E 3 are “measured” from the same point “ a ”, the potential across C must be E C = E 3 − E 1 = 1 V − 21 V = − 20 V |E C | = 20 V . 002 (part 2 of 2) 10.0 points If the battery is disconnected, how long does it take for the voltage across the capacitor to drop to a value of V ( t ) = E e , where E is the initial voltage across the capacitor? Correct answer: 91 μ s. Explanation: With the battery removed, the circuit is C I ℓ R 1 I r R 2 R 3 I ℓ I r R 4 ℓ r marout (ssm979) – hw16 – turner – (56725) 2 C R eq I eq where R ℓ = R 1 + R 3 = 7 Ω + 1 Ω = 8 Ω , R r = R 2 + R 4 = 9 Ω + 47 Ω = 56 Ω and R eq = parenleftbigg 1 R ℓ + 1 R r parenrightbigg − 1 = parenleftbigg 1 8 Ω + 1 56 Ω parenrightbigg − 1 = 7 Ω , so the time constant is τ ≡ R eq C = (7 Ω) (13 μ F) = 91 μ s . The capacitor discharges according to Q t Q = e − t/τ V ( t ) E = e − t/τ = 1 e − t τ = ln parenleftbigg 1 e parenrightbigg = − ln e t = τ (ln e ) = − (91 μ s) ( − 1) = 91 μ s . 003 10.0 points In the figure below the battery has an emf of 21 V and an internal resistance of 1 Ω . Assume there is a steady current flowing in the circuit. 5 μ F 5 Ω 9 Ω 1 Ω 21 V Find the charge on the 5 μ F capacitor. Correct answer: 63 μ C. Explanation: Let : R 1 = 5 Ω , R 2 = 9 Ω , r in = 1 Ω , V = 21 V , and C = 5 μ F . The equivalent resistance of the three resistors in series is R eq = R 1 + R 2 + r in = 5 Ω + 9 Ω + 1 Ω = 15 Ω , so the current in the circuit is I = V R eq , and the voltage across R 2 is V 2 = I R 2 = R 2 R eq V = 9 Ω 15 Ω (21 V) = 12 . 6 V . Since R 2 and C are parallel, the potential difference across each is the same, and the charge on the capacitor is Q = C V 2 = (5 μ F) (12 . 6 V) = 63 μ C ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 9

hw16-solutions - marout(ssm979 – hw16 – turner...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online