9) a) What is in volts, the voltage across the capacitor the instant switch S 1
ID: 2289987 • Letter: 9
Question
9)
a) What is in volts, the voltage across the capacitor the instant switch S1 was closed?
b) What is in volts, the voltage across the capacitor when it is fully charged?
c) What is the charge time constant for the circuit? Answer in units of seconds.
d) What is the discharge time constant for the circuit? Answer in units of seconds.
10)
a) What is the charge time constant for the circuit? Answer in units of seconds.
b) What is the discharge time constant for the circuit? Answer in units of seconds.
After fully discharging the circuit, we close S2 and then close S1. The capacitor will charge.
c) What is in micro-amps the current through R2 once the capacitor is fully charged?
d) What is in micro-amps the current through R1 once the capacitor is fully charged?
e) What is the voltage across the capacitor then?
11)
Explanation / Answer
1.)1.5 amperes 2.)From q=CV so 2*6= 12 micro Coulombs 3.)From E=1/2CV^2 so 1/2(2)(6^2) =36 micro joules.
For the work done, you can think of a differential deposition of charge on the capacitor plates (as the charge gets deposited on the capacitor). Please try and work this out (by considering the work done by the battery, the energy dissipated in the resistor and that stored in the capactor) yourself. Alternatively, you can say that the work done by the battery in the complete charging process is equal to the energy dissipated in the resistor plus the energy stored on the capacitor.
Lets calculate the expression for the work done by the battery: When S2 is closed, Kirchoff's Loop rule gives E=iR1+qC The current in the circuit is i=dq/dt. Multiplying both sides of this equation by dq=idt gives Edq=i2R1dt+qCdq The term on the left hand side is the work done by the battery (with an emf E and zero internal resistance) in time dt. As you can see, part of the work in getting a charge across to the circuit is dissipated as heat in the resistor (Joule heating, first term on the right hand side) and some of it is stored as electrostatic energy in the capacitor (second term on the right hand side). So for your problem, the work done by the battery is simply E?Q where ?Q is the total charge that leaves the battery in the charging period (from when the switch S1 is closed to steady state). The fourth question is: How much thermal heat is emitted from the resistor R1? (I have no idea about this) If a current i passes through a resistor R in time dt, the heat dissipated is equal to Vdq=(iR)(idt)=i2Rdt. The fifth question is: Keeping the switch S1 closed, the switch S2 is also closed. At a sufficiently long time after the switch S2 is closed, the circuit becomes steady again. How much charge is stored in the capacitor C long after the S2 is closed. (Am I right if the answer to this is the same as my answer in number two?) I am assuming that you are closing S2 after the steady state due to the left part of the circuit (not involving R2 and S2) has been achieved first. If so, the steady state conditions computed earlier are initial conditions when you close S2. Once you have closed S2, you now have two loops so you can either solve the circuit by assuming two currents in the two loops and writing the d.e. of the circuit (Kirchoff's Loop law really) or by using Thevenin's Theorem (if you know this, otherwise don't bother). What is the physics of the fifth problem then? Before you closed S2, capacitor C was an open circuit (due to steady state in the left half of the circuit). Now on closing S2, you provide the emf source E a conducting path (and also a path for the capacitor to discharge from Q=CE). How does the charge evolve in the circuit as a function of time
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