The charge entering the positive 2


The charge entering the positive terminal of an element is given by the expression q(t)=-12e^{-2t} mC. The power delivered to the element is p(t)=2.4e^{-3t} W. Compute the current in the element, the voltage across the element, and the energy delivered to the element in the time interval 0 < t < 100 ms.

Solution:

Show me the final answer↓

To find the current, we need to find the derivative of the charge equation. Remember that:

i=\dfrac{d(q(t))}{dt}

i=\dfrac{d(-12e^{-2t})}{dt}

(take the derivative)

i=24e^{-2t} mA or i=0.024e^{-2t} A

(First equation gives us mA while the second gives us A)

 

To find the voltage, remember that:

v(t)=\dfrac{p(t)}{i(t)}
 
v(t)=\dfrac{2.4e^{-3t}}{0.024e^{-2t}}
 
v(t)=100e^{-t}

 

 

The energy delivered to the element can be found by:

\,\displaystyle W=\int p(t)\,dt

\,\displaystyle W=\int^{0.1}_{0} (2.4e^{-3t})\,dt
 
W=-0.8e^{-3t}\Big|^{0.1}_{0}
 
W=0.207 J

 

Final Answers:

i=0.024e^{-2t} A
 
v(t)=100e^{-t}
 
W=0.207 J

 

This question can be found in Basic Engineering Circuit Analysis, 10th edition, chapter 1, question 1.11.

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