 # The current entering the positive terminal of a device is i(t) = 6e 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.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Paul Mitchell
a) ${\int }_{0}^{2}i\left(t\right)dt={\int }_{0}^{2}6\cdot {e}^{-2t}dt=6\cdot {\int }_{0}^{2}{e}^{-2t}dt=6\cdot \left(-\frac{1}{2}{e}^{-2t}\right){\mid }_{0}^{2}=$
$=6\cdot \left(-\frac{1}{2}\cdot \left({e}^{-2\cdot 2}-{e}^{-2\cdot 0}\right)\right)=-3\cdot \left({e}^{-4}-1\right)=2,945$ mC
b) $v\left(t\right)=10\cdot \frac{di}{dt}=10\cdot \frac{d}{dt}\cdot \left(6{e}^{-2t}\right)=10\cdot \left(-12{e}^{-2t}\right)=-120\cdot {e}^{-2t}$ mV

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