A 10.0 μFμF capacitor initially charged to 30.0 μCμC is discharged through a 1.30 kΩkΩ resistor. How long does it take to reduce the capacitor's charge to 15.0 μCμC ? Express your answer with the appropriate units. 0.052s

It takes approximately 0.009 seconds for the capacitor to discharge to half its original charge.

Explanation:

Recall that the capacitor discharges with an exponential decay associated with the time constant for the circuit () which in our case is;

(13 milliseconds)

Recall as well the expression for the charge in the capacitor (from it initial value , as it discharges via a resistor R:

So knowing that the capacitor started with a charge of 30 and reduced after a time "t" to 30 , and knowing from our first formula what the RC of the circuit is, we can solve for the time elapsed using the charge as function of time equation shown above:

In seconds this is approximately 0.009 seconds

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form