PHYS
19_InstSolManual_PDF_Part16

# 19_InstSolManual_PDF_Part16 - 19-16 Chapter 19 Solve(a P 5...

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Solve: (a) gives The loop rule applied to loop (1) gives: (b) The loop rule applied to loop (2) gives: The emf is 46.1 V and the polarity of the battery is opposite to what is shown in the figure in the problem; the terminal is adjacent to the resistor. 19.63. Set Up: The time constant is Solve: 19.64. Set Up: Solve: RC has units of 19.65. Set Up: The loop rule gives Solve: (a) The initial value of q is (b) Reflect: The larger the value of R, the smaller the initial current and the longer it takes the charge to decrease to of its initial value. 19.66. Set Up: The loop rule gives Just after the circuit is completed, After a long time the capacitor is fully charged and Solve: (a) so the voltage across the capacitor is zero. (b) so the voltage drop across the resistor is 500 V. (c) (d) (e) (a) so and the voltage across the capacitor is 500 V. (b) so the voltage across the resistor is zero. (c) (d) 19.67. Set Up: The time constant is Kirchhoff’s loop rule applied to the charging circuit gives at all times. The current and the charge on the capacitor as functions of time are given in Equations (19.17). Solve: (a) (b) when so Q final 5 E C
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• Spring '07
• Shoberg
• #, Electric charge, Electrical resistance, loop rule

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