Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 25, Problem 98P
To determine
The value of
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the capacitor discharged (V = 0, Q = 0) Should you close the switch? Why or why not?
Initially, for the circuit shown, the switch S is open and the capacitor is uncharged.The switch S is closed at time t = 0. In the following figure, when the time t is equal to 8.0 s, the charge on the capacitor, in μC, is closest to:
A) 1200
B) 940
C) 1400
D) 1600
E) 1900
Among the four different settings for values of \( R \) and \( C \) in the circuit shown below, which setting would result in the initially charged capacitor discharging more rapidly after the switch is closed?:
R = 3 Ω, C = 10 µF
R = 5 Ω, C = 3 µF
R = 3 Ω, C = 5 µF
R = 2 Ω, C = 5 µF
Chapter 25 Solutions
Physics for Scientists and Engineers
Ch. 25 - Prob. 1PCh. 25 - Prob. 2PCh. 25 - Prob. 3PCh. 25 - Prob. 4PCh. 25 - Prob. 5PCh. 25 - Prob. 6PCh. 25 - Prob. 7PCh. 25 - Prob. 8PCh. 25 - Prob. 9PCh. 25 - Prob. 10P
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- Referring to Figure CQ21.4, describe what happens to the light-bulb after the switch is closed. Assume the capacitor has a large capacitance and is initially uncharged. Also assume the light illuminates when connected directly across the battery terminals.arrow_forwardFor the circuit shown below, =20V , L = 4.0 mH, and R = 5.0 . After steady state is reached with S1 closed and S2 open, S2 is closed and immediately thereafter (at t = 0) S1 is opened. Determine (a) the current through L at t = 0, (b) the current through L at r = 4.0 x 10-4 s, and (c) the voltages across L and R at t = 4.0 x 10-4 s.arrow_forwardA capacitor with initial charge Q0 is connected across a resistor R at time t = 0. The separation between the plates of the capacitor changes as d = d0/(1 + t) for 0 t 1 s. Find an expression for the voltage drop across the capacitor as a function of time.arrow_forward
- In the figure V = 9.6 V, C1 = 9.8 µF, and C2 = C3 = 17 µF. Switch S is first thrown to the left side until capacitor 1 reaches equilibrium. Then the switch is thrown to the right. When equilibrium is again reached, how much charge is on capacitor 1?arrow_forwardWhile the switch is open, denote the energy stored in capacitor C1 with U0. After the switch has been closed for a long time, denote the energy stored in C1 with U1. Find the ratio U0/U1. Assume R1=1.87 ohm.arrow_forwardinitially, for the circuit shown, the switch S is open and the capacitor is uncharged. The switch S is closed at time t = 0. In the Figure, when the time t is equal to 8.0s, the charge on the capacitor, uC is closest to A) 240 B) 280 C) 190 D) 330 E) 380arrow_forward
- Consider the circuit shown in the diagram. Before the switch is closed, both capacitors are uncharged. V=9 VC1=3.0 μFC2=4.5 μFR1=80 ΩR2=20 ΩR3=70 Ωarrow_forwardWith the switch open, the potential difference across the capacitor shown is 10.0 V. After the switch is closed, how long will it take for the potential difference across the capacitor to decrease to 5.0 V?arrow_forwardConsider the circuit shown. Before the switch is closed, both capacitors are uncharged. V = 12.0 V, R1 = 90.0 Ω, R2 = 30.0 Ω, R3 = 60.0 Ω, C1 = 5.00 uf, C2 = 3.00 uf. Immediately after he switch is closed, what is the current in the circuit.arrow_forward
- The network shown in the figure is assembled with uncharged capacitors X, Y,and Z, with Cx= 7.0μF , CY= 7.0μF ,CZ= 6.0μF, and and open switches, S1 and S2. Apotential difference Vab = +120 V is applied between points a and b. After thenetwork is assembled, switch S1 is closed for a long time, but switch S2 is keptopen. Then switch S1 is opened and switch S2 is closed. What is the final voltageacross capacitor X? A) 94 VB) 87 VC) 79 VD) 71 VE) 63 Varrow_forwardA block of weakly-conducting material with conductivity sigma has two pieces of metalembedded inside, which form a capacitor with capacitance C.(a) Find the resistance R between the two pieces of metal.(b) At t = 0, the two pieces of metal have a potential difference V0, and at future times t not equal to 0the potential is V (t) = V0e-t/TFind Tarrow_forward27. Consider the circuit shown below. (A) Redraw the circuit step by step until only a single equivalent capacitor remains. Label each reduced capacitor using a symbol with subscripts. 24.0 nF 12V 18.0 nF I 36.0 nF 24.0 nF + 5.0 nk 6.0 nF 12.0 nF (B) Determine the equivalent capacitance of the circuitarrow_forward
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