A 27.0-µF capacitor and a 44.0-µF capacitor are charged by being connected acros
ID: 2026996 • Letter: A
Question
A 27.0-µF capacitor and a 44.0-µF capacitor are charged by being connected across separate 40.0-V batteries. Please fill in the blanks below.(a) Determine the resulting charge on each capacitor.
27.0-µF capacitor = ________________ µC
44.0-µF capacitor = ________________ µC
(b) The capacitors are then disconnected from their batteries and connected to each other, with each negative plate connected to the other positive plate. What is the final charge of each capacitor?
27.0-µF capacitor = ________________ µC
44.0-µF capacitor = ________________ µC
(c) What is the final potential difference across the 44.0-µF capacitor?
=______________________ V
Please help ASAP...Thank You in Advance.
Explanation / Answer
Given Data: Capacitances, C1 = 27.0 *10-6 F, C2 = 44 *10-6 F Potential difference, V1 = 40 V, V2 = 40 V Solution: a) Charge on each capacitor is, q1 = C1V1 = (27.0 *10-6 F) (40 V) = 1080 *10-6 C or 1080 C And q2 = C2V2 = (44.0 *10-6 F) (40 V) = 1760 *10-6 C or 1760 C ------------------------------------------------------------------------- b) When the capacitors are connected in series, then the potential difference on the capacitors is, V1 + V2 = 80 V ...... (1) Since, charge on the two capacitors is same, V1/C V1 /V2 = C2 /C1 = 44 F/ 27 F = 1.63 V1 = 1.63 V2 Taking equation (1), we get 1.63 V2 + V2 = 80 V V2 = 30.42 V And, V1 = 1.63 (30.42 V) = 49.5 V ---------------------------------------------------------------------------- Charge on each capacitor is, q1 = C1V1 = (27.0 *10-6 F) (49.5 V) = 1336.5*10-6 C or 1336.5 C And q2 = C2V2 = (44.0 *10-6 F) (30.42V) = 1338.5 *10-6 C or 1338.5 C ------------------------------------------------------------------------------- c) potential difference across the 50.0-µF capacitor is 30.42 V Given Data: Capacitances, C1 = 27.0 *10-6 F, C2 = 44 *10-6 F Potential difference, V1 = 40 V, V2 = 40 V Solution: a) Charge on each capacitor is, q1 = C1V1 = (27.0 *10-6 F) (40 V) = 1080 *10-6 C or 1080 C And q2 = C2V2 = (44.0 *10-6 F) (40 V) = 1760 *10-6 C or 1760 C ------------------------------------------------------------------------- b) When the capacitors are connected in series, then the potential difference on the capacitors is, V1 + V2 = 80 V ...... (1) Since, charge on the two capacitors is same, V1/C V1 /V2 = C2 /C1 = 44 F/ 27 F = 1.63 V1 = 1.63 V2 Taking equation (1), we get 1.63 V2 + V2 = 80 V V2 = 30.42 V And, V1 = 1.63 (30.42 V) = 49.5 V ---------------------------------------------------------------------------- Charge on each capacitor is, q1 = C1V1 = (27.0 *10-6 F) (49.5 V) = 1336.5*10-6 C or 1336.5 C And q2 = C2V2 = (44.0 *10-6 F) (30.42V) = 1338.5 *10-6 C or 1338.5 C ------------------------------------------------------------------------------- c) potential difference across the 50.0-µF capacitor is 30.42 V = (27.0 *10-6 F) (40 V) = 1080 *10-6 C or 1080 C And q2 = C2V2 = (44.0 *10-6 F) (40 V) = 1760 *10-6 C or 1760 C ------------------------------------------------------------------------- b) When the capacitors are connected in series, then the potential difference on the capacitors is, V1 + V2 = 80 V ...... (1) Since, charge on the two capacitors is same, V1/C V1 /V2 = C2 /C1 = 44 F/ 27 F = 1.63 V1 = 1.63 V2 Taking equation (1), we get 1.63 V2 + V2 = 80 V V2 = 30.42 V And, V1 = 1.63 (30.42 V) = 49.5 V ---------------------------------------------------------------------------- Charge on each capacitor is, q1 = C1V1 = (27.0 *10-6 F) (49.5 V) = 1336.5*10-6 C or 1336.5 C And q2 = C2V2 = (44.0 *10-6 F) (30.42V) = 1338.5 *10-6 C or 1338.5 C ------------------------------------------------------------------------------- c) potential difference across the 50.0-µF capacitor is 30.42 V = (44.0 *10-6 F) (40 V) = 1760 *10-6 C or 1760 C ------------------------------------------------------------------------- b) When the capacitors are connected in series, then the potential difference on the capacitors is, V1 + V2 = 80 V ...... (1) Since, charge on the two capacitors is same, V1/C V1 /V2 = C2 /C1 = 44 F/ 27 F = 1.63 V1 = 1.63 V2 Taking equation (1), we get 1.63 V2 + V2 = 80 V V2 = 30.42 V And, V1 = 1.63 (30.42 V) = 49.5 V ---------------------------------------------------------------------------- Charge on each capacitor is, q1 = C1V1 = (27.0 *10-6 F) (49.5 V) = 1336.5*10-6 C or 1336.5 C And q2 = C2V2 = (44.0 *10-6 F) (30.42V) = 1338.5 *10-6 C or 1338.5 C ------------------------------------------------------------------------------- c) potential difference across the 50.0-µF capacitor is 30.42 V = (27.0 *10-6 F) (49.5 V) = 1336.5*10-6 C or 1336.5 C And q2 = C2V2 = (44.0 *10-6 F) (30.42V) = 1338.5 *10-6 C or 1338.5 C ------------------------------------------------------------------------------- c) potential difference across the 50.0-µF capacitor is 30.42 V = (44.0 *10-6 F) (30.42V) = 1338.5 *10-6 C or 1338.5 C ------------------------------------------------------------------------------- c) potential difference across the 50.0-µF capacitor is 30.42 VRelated Questions
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