An initially uncharged 3.67 μF capacitor and a 8.01 k Ω resistor are connected in series to a 1.50 V battery that has negligible internal resistance. What is the initial current in the circuit, expressed in milliamperes?

The initial current in the circuit is 0.1873 mA

Explanation:

Given;

capacitance is given as 3.67 x 10⁻⁶ F

resistance is given as 8010 Ω

voltage across the circuit is 1.5 V

Since the capacitor is initially uncharged, the capacitive reactance is zero.

From ohms law;

Voltage across the circuit is directly proportional to the opposition to the flow of current.

V = IR

I = V/R

I = 1.5/8010

I = 0.0001873 A = 0.1873 mA

Therefore, the initial current in the circuit is 0.1873 mA

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