791_PartUniversity Physics Solution

791_PartUniversity Physics Solution - Direct-Current...

This preview shows pages 1–2. Sign up to view the full content.

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,

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.

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.

Page1 / 5

791_PartUniversity Physics Solution - Direct-Current...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online