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Sum

The electric current in a charging R−C circuit is given by i = i_{0}_{ }e^{−t}^{/RC} where i_{0}, R and C are constant parameters of the circuit and t is time. Find the rate of change of current at (a) t = 0, (b) t = RC, (c) t = 10 RC.

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#### Solution

Given: i = i_{0} e^{−t}^{/RC}

∴ Rate of change of current

\[= \frac{di}{dt}\]

\[= i_0 \left( \frac{- 1}{RC} \right) e^{- t/RC}\]

\[= \frac{- i_0}{RC} \times e^{- t/RC}\]

On applying the conditions given in the questions, we get:

(a) At \[t = 0, \frac{di}{dt} = \frac{- i_0}{RC} \times e^0 = \frac{- i_0}{RC}\]

(b) At \[t = RC, \frac{di}{dt} = \frac{- i_0}{RC} \times e^{- 1} = \frac{- i_0}{RCe}\]

(c) At \[t = 10 RC, \frac{dt}{di} = \frac{- i_0}{RC} \times e^{10}\] \[= \frac{- i_0}{RC e^{10}}\]

Concept: What is Physics?

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