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yulyashka [42]
3 years ago
8

What is the time constant of a series circuit where the capacitor is 0.330μF and the resistor is 10Ω ?

Physics
1 answer:
PtichkaEL [24]3 years ago
3 0

Answer:

\tau=3.3*10^{-6}s

Explanation:

Take at look to the picture I attached you, using Kirchhoff's current law we get:

C*\frac{dV}{dt}+\frac{V}{R}=0

This is a separable first order differential equation, let's solve it step by step:

Express the equation this way:

\frac{dV}{V}=-\frac{1}{RC}dt

integrate both sides, the left side will be integrated from an initial voltage v to a final voltage V, and the right side from an initial time 0 to a final time t:

\int\limits^V_v {\frac{dV}{V} } =-\int\limits^t_0 {\frac{1}{RC} } \, dt

Evaluating the integrals:

ln(\frac{V}{v})=e^{\frac{-t}{RC} }

natural logarithm to both sides in order to isolate V:

V(t)=ve^{-\frac{t}{RC} }

Where the term RC is called time constant and is given by:

\tau=R*C=10*(0.330*10^{-6})=3.3*10^{-6}s

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