It will take 2.3sec to charge 90%.
The charge on the capacitor is changing in time as follows:
q(t)=qo (1−e−tRC)
Here
• qo is the maximum charge on the capacitor.
• R=50000 Ω is the resistor value.
• C=2⋅10−6 F is the capacitance of the capacitor.
Solving for the time, we obtain:
t=−RCln(1−qq0)
Here
• q=0.9qo is the charge on the capacitor at the moment t;
The Computation yields:
t=−5⋅[tex]10^{4}[/tex]Ω⋅2⋅[tex]10^{-6}[/tex] F⋅ln(1−0.9q0)
≈0.23 s
C or Q may be equal. Thus, the charge needed to raise a conductor’s potential through unity is numerically equivalent to the conductor’s capacitance.
There is no current flowing across the circuit when a capacitor is fully charged. This is true because the voltage source’s potential difference across the capacitor is also equal.
In other words, the charging current reaches zero and the capacitor voltage equals the source voltage.
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