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Chapter 8: Problem 2

The current in a circuit, \(i(t)\), is given by $$ i(t)=25 \mathrm{e}^{-0.2 t} \quad t \geq 0 $$ (a) State the current when \(t=0\). (b) Calculate the value of the current when \(t=2\) (c) Calculate the time when the value of the current is \(12.5\).

Short Answer

Expert verified
Answer: The current is 25 A when t = 0, approximately 18.34 A when t = 2, and the current is 12.5 A at approximately 3.47 seconds.

Step by step solution

01

(a) Current at t = 0

To find the current when t = 0, we need to plug t = 0 into the given current function: $$ i(t) = 25 e^{-0.2 t} $$ When \(t = 0\): $$ i(0) = 25 e^{-0.2 \times 0} = 25 e^0 = 25(1) = 25 $$ So the current when \(t = 0\) is 25 A.
02

(b) Current at t = 2

To find the current when t = 2, we need to plug t = 2 into the given current function: $$ i(t) = 25 e^{-0.2 t} $$ When \(t = 2\): $$ i(2) = 25 e^{-0.2 \times 2} = 25 e^{-0.4} \approx 18.34 $$ So the current when \(t = 2\) is approximately 18.34 A.
03

(c) Time when current is 12.5

To find the time when the current is 12.5, we need to set the current function equal to 12.5 and solve for t: $$ 12.5 = 25 e^{-0.2 t} $$ First, divide both sides by 25: $$ 0.5 = e^{-0.2 t} $$ Now take the natural logarithm of both sides: $$ \ln(0.5) = -0.2 t $$ Finally, divide by -0.2 to get t: $$ t = \frac{\ln(0.5)}{-0.2} \approx 3.47 $$ So, the time when the current is 12.5 A is approximately 3.47 seconds.

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