The voltage v(t) across a device and the current i(t) through it are v(t) = 20 sin (4t) V and i(t) = 10(1 + e−2t ) mA Calculate: (a) the total charge in the device at t = 1 s, q(0) = 0.

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madeleinejames20 · Accepted Answer

To calculate the total charge in the device at t = 1 s, we need to integrate the current function from t = 0 to t = 1.The current function is given by:i(t)=10(1+e−2t)mATo integrate this function, we&#039;ll use the integral symbol and the appropriate limits of integration:q(t)=∫0ti(t)dtSubstituting the given current function into the integral, we have:q(t)=∫0t10(1+e−2t)dtTo solve this integral, we&#039;ll first integrate each term separately.For the term 10 which is a constant, we can pull it out of the integral:q(t)=10∫0t(1+e−2t)dtThe integral of 1 with respect to t is t:q(t)=10(t+∫0te−2tdt)Now, we need to find the integral of e^(-2t) with respect to t. To do this, we can use the substitution method. Let&#039;s substitute u = -2t and find du:du=ddt(−2t)dtdu=−2dtRearranging the equation, we have dt = -(1/2) du.Substituting these values into the integral, we get:q(t)=10(t+∫0teu(−12)du)Simplifying further:q(t)=10(t−12∫0teudu)Now, integrating e^u with respect to u gives us e^u:q(t)=10(t−12[eu]0t)Substituting back u = -2t:q(t)=10(t−12[e−2t]0t)Now, let&#039;s substitute the limits of integration:q(t)=10(t−12[e−2t−e0])Simplifying further, we have:q(t)=10(t−12[e−2t−1])Finally, we can substitute t = 1 s into the equation to find the total charge at t = 1 s:q(1)=10(1−12[e−2])Calculating the exponential term:q(1)=10(1−12[0.1353])q(1)=10(1−0.0677)q(1)=10×0.9323q(1)=9.323mCTherefore, the total charge in the device at t = 1 s is approximately

The voltage v(t) across a device and the

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