Josh Sizemore
2021-12-22
Answered

The current entering the positive terminal of a device is $i\left(t\right)=6{e}^{-2t}$ mA and the voltage across the device is v(t) = 10di/dt V. (a) Find the charge delivered to the device between t = 0 and t = 2 s. (b) Calculate the power absorbed. (c) Determine the energy absorbed in 3 s.

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Paul Mitchell

Answered 2021-12-23
Author has **40** answers

a) ${\int}_{0}^{2}i\left(t\right)dt={\int}_{0}^{2}6\cdot {e}^{-2t}dt=6\cdot {\int}_{0}^{2}{e}^{-2t}dt=6\cdot (-\frac{1}{2}{e}^{-2t}){\mid}_{0}^{2}=$

$=6\cdot (-\frac{1}{2}\cdot ({e}^{-2\cdot 2}-{e}^{-2\cdot 0}))=-3\cdot ({e}^{-4}-1)=2,945$ mC

b)$v\left(t\right)=10\cdot \frac{di}{dt}=10\cdot \frac{d}{dt}\cdot \left(6{e}^{-2t}\right)=10\cdot (-12{e}^{-2t})=-120\cdot {e}^{-2t}$ mV

$p\left(t\right)=v\left(t\right)\cdot i\left(t\right)=-120\cdot {e}^{-2t}\cdot 6{e}^{-2t}=-720\cdot {e}^{-4t}\text{}\mu W$

c)$W={\int}_{0}^{3}p\left(t\right)dt=-720\cdot {\int}_{0}^{3}{e}^{-4t}dt=-720\cdot (-\frac{1}{4}{e}^{-4t}){\mid}_{0}^{3}=$

$=180{e}^{-4t}{\mid}_{0}^{3}=180({e}^{-12}-{e}^{0})=180({e}^{-12}-1)=-180\text{}\mu J$

b)

c)

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