After the capacitor is fully charged, you switch the switch to position b at time t = 0 seconds. The voltmeter initially reads 4.5 V. After how much elapsed time (in seconds) would you expect the voltmeter to read 1/2 its initial value?

0.693RC where R is resistance in the circuit and C is capacitance of the capacitor

Explanation:

v=V×e^(-t/RC)

V:Initial Voltage during discharge

t: time

R: resistance in the discharge circuit

C: Capacitance of the capacitor

Consider the circuit in the attachment

For v=0.5V

0.5V=V×e^(-t/RC)

t=0.693RC

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

a.) travels faster in solids because the particles are closer together.

air is a gas water is a liquid