Here is the link. I am totally lost please help me.... 9. Now click on the Multi
ID: 1795254 • Letter: H
Question
Here is the link.
I am totally lost please help me....
9. Now click on the Multiple Capacitors tab at the top of the simulation, and explore how capacitance, plate charge, voltage, and stored energy vary with different configurations of capacitors 10. Set the battery voltage to some constant value like 1 V throughout your exploration. For 2 capacitors in series, use the voltmeter to measure the voltage across the battery, and across each capacitor. Change the capacitance of one of the capacitors using the slider bars. a. b. Verify for Capacitors in Series: Voltage adds to equal the voltage of the battery ii. i. Charge is the same on the plates (you'll have to calculate how much charge is on each independently using Q- CV) iii. Cequivalent (1/C 1/C2)*1 For 2 capacitors in parallel, use the voltmeter to measure the voltage across the battery, and across each capacitor. Change the capacitance of one of the capacitors using the slider bars. Verify for Capacitors in Parallel: c. d. i. Voltage across each capacitor stays the same ii. Charge on each plate adds to the total charge ii. C equivalent Ci+2 11. Select 2 Capacitors in Series + 1 in Parallel, and develop an experiment that demonstrates the expected value for equivalent capacitance. Make a sketch of your circuit, and record all values. Run the experiment with at least three trials, with different values of capacitances for each of the components. Verify that charge is the same on components in series, and voltage is the same across components in parallel.Explanation / Answer
9. form the simulation we can conclude
Capacitances add algebriacally in parallel, and inverse of inverse addition in series
Plate chaarge is equal for capacitors in series, and adds algebriacally in parallel
Voltage is added algebriacally for capacitors in series, and remains same for capacitors in parallel
Stored energy always adds up algebriacally
10. V = 1 V
a. 2 capacitors in series
V = 1 V ( VOltage across battery)
Voltage across capacitor 1 = V1, across capacitor 2 = V2
hence V1 = q1/C1
V2 = q2/C2
but
q1 = q2 = q
and V1 + V2 = V
hence
q(1/C1 + 1/C2) = V
q = VC1C2/(C1 + C2) = Ceq*V
hence
V1 = VC2/(C1 + C2)
V2 = VC1/(C1 + C2)
b. it can be verified
i. V = V1 + V2
ii. q1 = q2 = q
iii. Ceq = C1C2/(C1 + C2) (from perevious part)
c. for 2 capacitors in parallel
Voltage drop across each = V
q1 = C1V
q2 = C2V
(q1 + q2) = (C1 + C2)V = Ceff V
d. it is verified frpm the previous part
i. V1 = V2 = V
ii. q1 + q2 = Q
iii. Ceff = C1 + C2