A proposed house is to be located on a site of 150 ft times 100 ft. Suppose that
ID: 1421641 • Letter: A
Question
A proposed house is to be located on a site of 150 ft times 100 ft. Suppose that the site needs to be raised 2.0 ft. higher. the soil will be compacted to a bulk unit weight of 112.5 pcf and a water content of 12.5% when it is placed on site. There are three borrow pits available, designated A, B, and C. the bulk unit weight and corresponding water content of the soil in each pit and the cost per cubic yard for the four different soils are shown as follows: Borrow pit designation Bulk unit weight (pcf) Water Content (%) Cost/yd3Select the pit from which the fill should be bought so that the cost is minimized. Ignore factors pertaining to labor cost involved in compaction and changing the water content.Explanation / Answer
Hi,
To solve this problem we should remember that the charge and the voltage are related by the capacitance as shown:
Q = CV ; where Q is the charge, V is the electric potential and C is the capacitance.
(a) In this case, as S2 is opened and S1 is closed the charge will flow only to C1, so the adquired by the C1 is:
Q1 = C1V = (2*10-6 F)(22 V) = 44*10-6 C
(b) For this part, as the battery is cut from the circuit we have now a system form by the two capacitors. Once C1 and C2 are connected, a new potential difference is established. That voltage is the responsible for charging C2.
To find that potential and the new charge in each capacitor we have to do a balance of charge, at the beginning and at the end.
At the beginning
Qo = Qo1 + Qo2; but Qo2 = 0 (beacuse it was not connected to the battery, and Qo1 = C1V)
Qo = C1V
At the end
Qf = Qf1 + Qf2 = C1Vf + C2Vf = (C1 + C2)Vf
The charge can neither be created nor destroyed, so:
Qo = Qf ::::::::: Vf = V*C1/(C1+C2) = (22 V)*[ 2/(2+1) ] = 14.7 V
So the charge in each capacitor is:
Qf1 = VfC1 = (14.7 V)*(2*10-6 F) = 29.4*10-6 C
Qf2= VfC2 = (14.7 V)*(1*10-6 F) = 14.7*10-6 C
I hope it helps.