In Figure P28.67, suppose the switch has been closed for a length of time sufficiently long for the capacitor to become fully charged. ( = 9.40 V, r1 = 10 kA?©, and r2 = 16 kA?©.) (a) Find the steady-state current in each resistor. I1 = A?µA I2 = A?µA I3-kA?© = A?µA (b) Find the charge Q on the capacitor. A?µC (c) The switch is opened at t = 0. Write an equation for the current IR2 in R2 as a function of time. (304 A?µA) et/(0.190 s) (362 A?µA) eAcˆ’t/(0.190 s) (304 A?µA) eAcˆ’t/(0.190 s) (362 A?µA) et/(0.190 s) (d) Find the time that it takes for the charge on the capacitor to fall to one-fifth its initial value. ms
In Figure P28.67, suppose the switch has been closed for a length of time sufficiently long for the capacitor to become fully charged. ( = 9.40 V, r1 = 10 kA?©, and r2 = 16 kA?©.) (a) Find the steady-state current in each resistor. I1 = A?µA I2 = A?µA I3-kA?© = A?µA (b) Find the charge Q on the capacitor. A?µC (c) The switch is opened at t = 0. Write an equation for the current IR2 in R2 as a function of time. (304 A?µA) et/(0.190 s) (362 A?µA) eAcˆ’t/(0.190 s) (304 A?µA) eAcˆ’t/(0.190 s) (362 A?µA) et/(0.190 s) (d) Find the time that it takes for the charge on the capacitor to fall to one-fifth its initial value. ms
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter21: Current And Direct Current Circuits
Section: Chapter Questions
Problem 34P: Two 1.50-V batterieswith their positive terminals in the same directionare inserted in series into a...
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In Figure P28.67, suppose the switch has been closed for a length of time sufficiently long for the capacitor to become fully charged. ( = 9.40 V, r1 = 10 kA?©, and r2 = 16 kA?©.) (a) Find the steady-state current in each resistor. I1 = A?µA I2 = A?µA I3-kA?© = A?µA (b) Find the charge Q on the capacitor. A?µC (c) The switch is opened at t = 0. Write an equation for the current IR2 in R2 as a function of time. (304 A?µA) et/(0.190 s) (362 A?µA) eAcˆ’t/(0.190 s) (304 A?µA) eAcˆ’t/(0.190 s) (362 A?µA) et/(0.190 s) (d) Find the time that it takes for the charge on the capacitor to fall to one-fifth its initial value. ms
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