Question

The current entering the positive terminal of a device is i(t)= 6e^-2t mA and the voltage across the device is v(t)= 10di/dtV.

Answer 1

Answer:

a) 2,945 mC

b) P(t) = -720*e^(-4t) uW

c) -180 uJ

Explanation:

Given:

                           i (t) = 6*e^(-2*t)

                           v (t) = 10*di / dt

Find:

( 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.

Solution:

-  The amount of charge Q delivered can be determined by:                      

                                       dQ = i(t) . dt

                  $Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt$

- Integrate and evaluate the on the interval:

                   $= 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C$

- The power can be calculated by using v(t) and i(t) as follows:

                 v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt

                 v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV

                 P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)

                 P(t) = -720*e^(-4t) uW

- The amount of energy W absorbed can be evaluated using P(t) as follows:

                 $W = \int\limits^3_0 {P(t)} \, dt = \int\limits^2_0 {-720*e^(-4t)} \, dt = -720*\int\limits^2_0 {e^(-4t)} \, dt$

- Integrate and evaluate the on the interval:

                  $W = -180*e^-4t = - 180*( 1 / e^12 - 1) = -180uJ$