791_PartUniversity Physics Solution

791_PartUniversity Physics Solution - Direct-Current...

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Direct-Current Circuits 26-33 (b) SET UP: For discharging, U = Q 2 /2 C = [ Q 0 (1 – e t / RC )] 2 /2 C = U max (1 – e t / RC ) 2 EXECUTE: To reach 1/ e of the maximum energy, U ma x / e = U max (1 – e t / RC ) 2 and t = 1 ln 1 RC e ⎛⎞ −− ⎜⎟ ⎝⎠ . EVALUATE: The time to reach 1/ e of the maximum energy is not the same as the time to discharge to 1/ e of the maximum energy. 26.87. IDENTIFY and SET UP: For parts (a) and (b) evaluate the integrals as specified in the problem. The current as a function of time is given by Eq.(26.13) / . tRC ie R = E The energy stored in the capacitor is given by 2 /2 . QC EXECUTE: (a) Pi = E The total energy supplied by the battery is () ( ) 2/2 / 2 0 00 0 // . tR C C Pdt idt R e dt R RCe C ∞∞ ⎡⎤ == = = ⎣⎦ ∫∫ EE E E (b) 2 PiR = The total energy dissipated in the resistor is 22 2 / 2 2 / 2 1 2 0 0 / 2 . P d t i R d e d R C e C =− = E (c) The final charge on the capacitor is . = E The energy stored is ( ) 1 2 /2 . UQ C C E The final energy stored in the capacitor ( ) 2 1 2 C = E total energy supplied by the battery ( ) 2 C E – energy dissipated in the resistor ( ) 2 1 2 C E (d) EVALUATE: 1 2 of the energy supplied by the battery is stored in the capacitor. This fraction is independent of R . The other 1 2 of the energy supplied by the battery is dissipated in the resistor. When R is small the current initially is large but dies away quickly. When R is large the current initially is small but lasts longer. 26.88. IDENTIFY: 0 E Pdt = . The energy stored in a capacitor is 2 Uq C = . SET UP: / 0 C Q RC EXECUTE: 0 Q RC gives 2 / 0 2 C Q e RC and 2 2 0 0 0 . QQ R C Q E ed t U RC RC C = = EVALUATE: Increasing the energy stored in the capacitor increases current through the resistor as the capacitor discharges. 26.89. IDENTIFY and SET UP: EXECUTE: (a) Using Kirchhoff’s Rules on the circuit we find: Left loop: 12 92 140 210 55 0 147 140 210 0. II + = = Right loop: 32 2 3 57 35 210 55 0 112 210 35 0. I I + = −= Junction rule: 123 0. III −+= Solving for the three currents we have: 1 0.300 A, I = 2 0.500 A, I = 3 0.200 A. I = (b) Leaving only the 92-V battery in the circuit: Left loop: 92 140 210 0. = Right loop: 35 210 0. Junction rule: 0. II I Solving for the three currents: 1 0.541A, I = 2 0.077 A, I = 3 0.464 A. I (c) Leaving only the 57-V battery in the circuit: Left loop: 140 210 0. += Right loop: 57 35 210 0. Junction rule: 0. Solving for the three currents: 1 0.287 A, I 2 0.192 A, I = 3 0.480 A. I = (d) Leaving only the 55-V battery in the circuit: Left loop: 55 140 210 0. = Right loop: 55 35 210 0. Junction rule: 12 3 0. Solving for the three currents: 1 0.046 A, I = 2 0.231A, I = 3 0.185 A. I = (e) If we sum the currents from the previous three parts we find: 1 0.300 A,
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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791_PartUniversity Physics Solution - Direct-Current...

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