A capacitor of 20.0 μ F and a resistor of 90.0 Ω are quickly connected in series to a battery of 8.00 V. What is the charge Q on the capacitor 0.00200 s after the connection is made? Q =

Q=107μC

Explanation:

Given an RC series circuit

Input voltage V=8V

Resistor =90Ω

And the capacitor=20 μF

Charge Q? After t=0.002

In an RC circuit,

The capacitor voltage is given as

Vc=V(1-e^-t/τ )

Where τ is time constant

τ =RC

τ =90×20×10^-6

τ=0.0018s

Therefore,

Vc=V(1-e^-t/τ)

Given that V=8V and t=0.002

Then,

Vc=8(1-e^-0.002/0.0018)

Vc=8(1-e^-1.111)

Vc=8(1-0.329)

Vc=8×0.671

Vc=5.37Volts

Then the voltage across the capacitor is 5.37V

From the relation

Q=CV, we can calculate the charge

Q=20×10^-6×5.37

Q=1.07×10^-4 C

Q=107μC

answer: strength i guess

explanation:

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge