For a series LRC circuit, if L=30H, R= 0.05 ohms and C= 5.8F, what is the time required for the amplitude of the charge stored by the capacitor to be halved?
We know that
Charge
q=q0(1−e−tRC)q=q_0(1-e^{\frac{-t}{RC}})q=q0(1−eRC−t)
q=q0(1−e−t0.05×5.8)q=q_0(1-e^{\frac{-t}{0.05\times5.8}})q=q0(1−e0.05×5.8−t)
q02=q0(1−e−t0.05×5.8)\frac{q_0}{2}=q_0(1-e^{\frac{-t}{0.05\times5.8}})2q0=q0(1−e0.05×5.8−t)
12=e−t0.05×5.8\frac{1}{2}=e^{-\frac{t}{0.05\times5.8}}21=e−0.05×5.8t
ln2=t0.05×5.8ln2=\frac{t}{0.05\times5.8}ln2=0.05×5.8t
t=0.29×ln2=0.202sect=0.29\times ln2=0.202sect=0.29×ln2=0.202sec
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