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raketka [301]
3 years ago
9

A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti

me constant for the process is about:
Physics
1 answer:
vichka [17]3 years ago
5 0

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

q(t) = e^{-\frac{t}{(R*C)}}

Here,

q = Charge

t = Time

R = Resistance

C = Capacitance

When the charge reach its half value it has passed 10ms, then the equation is,

\frac{1}{2}*q_{final} = e^{-\frac{0.01}{(R*C)}}

Ln(\frac{1}{2}) = -\frac{0.01}{RC}

- RC = \frac{0.01}{Ln(1/2)}

RC = 0.014s

We know that RC is equal to the time constant, then

T = RC  = 0.014s = 14ms

Therefore the time constant for the process is about 14ms

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wolverine [178]
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v = 11.5 m/s, final speed
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