PHYS 6B Lecture Notes - Lecture 12: Resistance 2, Rc Time Constant, Internal Resistance

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27 Aug 2016

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A resistor in a series circuit can have a greater resistance and greater power. So a circuit with a bigger resistor has a smaller amount of current. V1=ir1 and v2=ir2 but r2>r1 so v2>v1 so energy lost by a single charge in a resistor is times by the current. Power will be larger in r2 than in r1 because it is the energy per charge that is different for r2 and r1 but the current will be the same for both of the resistors. Ex: running a race that has two muddy paths. One is longer than the other so it will take a long time, take a lot of energy (r2) The short one is cumbersome but does not take a lot of energy (r1) When a resistor is used in a circuit, it warms up. The resistance of a given resistor depends on dimensions and resistivity of the material.

Related Questions

An ideal voltmeter connected across a certain 9-V fresh battery reads 9.30 V, and an ideal ammeter briefly connected across the same battery reads 3.70 A. We say the battery has an open-circuit voltage of 9.30 V and a short-circuit current of 3.70 A. Model the battery as a source of emf ε in series with an internal resistance r. Determine both (a) ε and (b) r. An experimenter connects two of these identical batteries together as shown in the figure below. Find (c) the open-circuit voltage and (d) the short-circuit current of the pair of connected batteries. (e) The experimenter connects a 12.0-Ω resistor between the exposed terminals of the connected batteries Find the current in the resistor. (f) Find the power delivered to the resistor. (g) The experimenter connects a second identical resistor in parallel with the first. Find the power delivered to each resistor. (h) Because the same pair of batteries is connected across both resistors as was connected across the single resistor, why is the power in part (g) not the same as that in part (f)?

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The power P delivered to each resistor is the product of thecurrent squared (I^2) and the corresponding resistance R, or P =I^2*R. The resistances are known, and Ohm's law can be used to findthe current. Ohm's law states that the current in the circuit(which is also the current through each of the resistors) equalsthe voltage V of the battery divided by the equivalent resistanceRs of the two resistors: I = V/Rs. Since the resistors areconnected in series, we can obtain the equivalent resistance byadding the resistances.

Three resistors, 3 O, 4.5 O, and 5.5 O, are connected in seriesacross a 12-V battery. Find the power delivered to eachresistor.

This set of problems is designed to help you prepare for the lab oncapacitors. Here are some things you should know.
The charge on a capacitor is proportional to the voltage. Theconstant of proportionality is the capacitance:
q=VC.
The simplest type of capacitor is a parallel plate capacitorconsisting of twoparallel conducting plates...DUH. If the plateseach have area A and
are separated by a distance d, the capacitance will be
C=Ke0A/d,
where e0 is the permittivity of free space (you measured that lastweek) and K is the dielectric constant of what ever material isbetween the plates. If the space between the plates is filled withvacuum, the dielectric constant is exactly 1. Air is pretty darnclose to 1.
When a capacitor is charged in a series circuit with a resistor,the charge builds up with time. Consider the circuit below.

When the switch, S, is thrown to position A, the battery, withvoltage V0, will supply a current through the resistor, R, thatcharges the capacitor C. The voltage across the capacitor at anygiven time is given by
V=V0(1-e-t/RC).
The product RC is known as the time constant. After waiting a timeequal to 1 time constant, the voltage on the capacitor is(1-e-1)=0.632 of V0. After 5RC the capacitor is nearly fullycharged and has a voltage of 0.993V0. Eventually, the voltagebuilds up to V0.
When the switch is thrown to position B, the capacitor willdischarge through the resistor. The voltage as a fuction of timewill be given by
V=V0(e-t/RC).
After a long enough time, the capacitor will fully discharge theresistor.

1. [1pt]
A parallel plate capacitor has plates with length and width of 5.7cm and are separated by 7.0 mm. The capacitance is measured to be1.479Ã10-11 F. What is the dielectric constant of the materialbetween the plates?

Correct, computer gets: 3.60

2. [1pt]
Consider a series circuit like the one shown above with a resistorof R=80. kilo-ohms and a capacitor of C=8.0 uF. What is the timeconstant of the circuit?

Correct, computer gets: 0.64 s

Hint: Don't forget to convert to Farads and ohms from uF andkilo-ohms.
3. [1pt]
If the battery has a voltage of 12 V, what is the voltage acrossthe capacitor after the switch has been in position A for 2 timeconstants?

Correct, computer gets: 10.38 V
4. [1pt]
What is the final charge on the capacitor after a long time?

Correct, computer gets: 9.60E-05 C
5. [1pt]
Now let's say that the capacitor has been fully charged and wethrow the switch to position B. How long will it take for thevoltage on the capacitor to be 0.891 V?

6. [1pt] What is the charge on the capacitor at this time?

Answer: ____________

PHYS 6B Lecture Notes - Lecture 12: Resistance 2, Rc Time Constant, Internal Resistance

PHYS 6B Full Course Notes

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