Solveeit Logo
Login

Question

Question: The electric current in a discharging R-C circuit is given by \( i = {i_ \circ }{e^{ - t/RC}} \) whe...

The electric current in a discharging R-C circuit is given by i=i∘e−t/RCi = {i_ \circ }{e^{ - t/RC}} where i∘,R&C{i_ \circ },R\& C are constant parameters and tt is in time. Let i∘=2.00A,R=6.00×105Ω,C=0.5μF{i_ \circ } = 2.00A,R = 6.00 \times {10^5}\Omega ,C = 0.5\mu F .
Find:
(a)\left( a \right) Current at t=0.3sec⁡t = 0.3\sec
(b)\left( b \right) Rate of change of current at t=0.3sec⁡t = 0.3\sec .

Explanation

Solution

Since we have the equation of electric current in a discharging R-C circuit is given by i=i∘e−t/RCi = {i_ \circ }{e^{ - t/RC}} . So by substituting the values and solving for the value of current we will get the answer. For the next question, First of all, we will differentiate the current as we know that the rate of change can be represented in the form of differentiation. And then solving for the specified time, we will get the answer for this.

Complete Step By Step Answer:
So we have the equation given by i=i∘e−t/RCi = {i_ \circ }{e^{ - t/RC}}
From the question, we have the values i∘=2.00A,R=6.00×105Ω,C=0.5μF{i_ \circ } = 2.00A,R = 6.00 \times {10^5}\Omega ,C = 0.5\mu F
On substituting the value of i∘,R&C{i_ \circ },R\& C in the equation given in the question, we get
⇒i=2.0e−t/0.3\Rightarrow i = 2.0{e^{ - t/0.3}}
So according to the question current at t=0.3sec⁡t = 0.3\sec , will be given by substituting the values in the above equation of current
⇒i=2.0e−0.3/0.3\Rightarrow i = 2.0{e^{ - 0.3/0.3}}
And on solving the above equation, we get
⇒i=2.0e−1\Rightarrow i = 2.0{e^{ - 1}}
And it can be written as i=2ei = \dfrac{2}{e}
Hence, the current at t=0.3sec⁡t = 0.3\sec will be equal to 2e\dfrac{2}{e} .
Now differentiating the question equation, we will get the equation as
⇒didt=−i∘RC⋅e−t/RC\Rightarrow \dfrac{{di}}{{dt}} = \dfrac{{ - {i_ \circ }}}{{RC}} \cdot {e^{ - t/RC}}
So when t=0.3sec⁡t = 0.3\sec on substituting the value of current we will get the equation as
⇒didt=20.30⋅e−0.3/0.3\Rightarrow \dfrac{{di}}{{dt}} = \dfrac{2}{{0.30}} \cdot {e^{ - 0.3/0.3}}
And on solving the above equation we will get the equation as
⇒didt=−203eA/s\Rightarrow \dfrac{{di}}{{dt}} = \dfrac{{ - 20}}{{3e}}A/s
Hence, the rate of change of current at t=0.3sec⁡t = 0.3\sec will be equal to −203eA/s\dfrac{{ - 20}}{{3e}}A/s .

Note:
For solving this type of question we have to just know the calculation and a little bit of differentiation. Also, we should know whenever there is the term rate of change given then we should always think that it is the derivative which is given to us.