Physics9C_A_HW7_Soln

Physics9C_A_HW7_Soln - DIRECT-CURRENT CIRCUITS 26.38 26 IDENTIFY An uncharged capacitor is placed into a circuit Apply the loop rule at each time

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26-1 D IRECT -C URRENT C IRCUITS 26.38. IDENTIFY: An uncharged capacitor is placed into a circuit. Apply the loop rule at each time. SET UP: The voltage across a capacitor is / C Vq C = . EXECUTE: (a) At the instant the circuit is completed, there is no voltage over the capacitor, since it has no charge stored. (b) Since 0 C V = , the full battery voltage appears across the resistor 125 V. R V == E (c) There is no charge on the capacitor. (d) The current through the resistor is total 125 V 0.0167 A. 7500 i R = E (e) After a long time has passed the full battery voltage is across the capacitor and 0 i = . The voltage across the capacitor balances the emf: 125 V. C V = The voltage across the resister is zero. The capacitor’s charge is 64 (4.60 10 F) (125 V) 5.75 10 C. C qC V −− × The current in the circuit is zero. EVALUATE: The current in the circuit starts at 0.0167 A and decays to zero. The charge on the capacitor starts at zero and rises to 4 5.75 10 C q . 26.41. IDENTIFY: The capacitors, which are in parallel, will discharge exponentially through the resistors.
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This note was uploaded on 09/22/2011 for the course PHY 09C 9c taught by Professor Zhu during the Spring '11 term at UC Davis.

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Physics9C_A_HW7_Soln - DIRECT-CURRENT CIRCUITS 26.38 26 IDENTIFY An uncharged capacitor is placed into a circuit Apply the loop rule at each time

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