Answer to Question #259236 in Electric Circuits for Rex

Question

Determine the current flowing through an element if the charge flow is given by a. q(t)= (3e-t -5e-2t) nC b. q(t)= 10sin120πt pC c. q(t)= 20e-4tcos50t µC

Answer 1

If the charge flow is given as a function of time, i.e. q(t), we determine the current flowing through an element by:

$I = \frac {dq(t)} {dt}$

where I = current

(a) For the charge flow given by: q(t)= (3e-t -5e-2t) nC;

$I = \frac {dq(t)} {dt} \\ I = \frac {d(3e-t -5e-2t) nC} {dt} \\ I = \frac {d} {dt} (3e-t -5e-2t) nC \\$

differentiating q(t) = (3e-t -5e-2t) nC with respect to time (t),

$I = \frac {d} {dt} (3e) - \frac {d} {dt} (t) - \frac {d} {dt} (5e)- \frac {d} {dt} (2t) \ nC \\ I = (0 - 1 - 0 - 2) nC \\ I = -3 nC \\ or \ I = -3 \times 10^{-9} \ C$

(b) For the charge flow given by: q(t) = 10sin120πt pC

$I = \frac {dq(t)} {dt} \\ I = \frac {d(10 \sin120\pi t) \ pC} {dt}$

differentiating q(t) = 10sin120πt pC with respect to time (t),

$I = \frac {d} {dt} (10 \sin 120\pi t) pC \\ let \ w = 120\pi t \\ u= \sin w \\ and, \ q = 10u \\ I = \frac {dq} {dt} = \frac {dw} {dt} \times \frac {du} {dw} \times \frac {dq} {du} \\ w = 120 \pi t, \ \frac {dw} {dt} = \frac {d} {dt} (120\pi t) = 120 \pi \\ u = \sin w, \frac{du}{dw} = \frac {d} {dw} (\sin w) = \cos w \\ q = 10 u, \frac {dq} {du} = \frac {d} {du} (10u) = 10 \\ I = \frac {dq} {dt} = 120 \pi \times \cos w \times 10 \ pC \\ I = 1200 \pi \cos 120 \pi t \ pC \\ or \\ I = 1200 \pi \cos 120 \pi t \times 10^{-12} \ C \\ I = 1.2 \pi \cos 120 \pi t \times 10^{-9} \ C \\ I = 1.2 \pi \cos 120 \pi t \ nC$

(c) For the charge flow given by: q(t)= 20e-4tcos50t µC

$I = \frac {dq(t)} {dt} \\ I = \frac {d(20e-4tcos50t) \ µC} {dt} \\$

differentiating q(t) = 20e-4tcos50t µC with respect to time (t),

$I = \frac {d} {dt} (20e-4t\cos50t) \ µC \\ I = \frac {d} {dt} (20e) - \frac {d} {dt} 4t\cos50t \\ \frac {d} {dt} (20e) = 0 \\ \therefore I = - \frac {d} {dt} (4t\cos50t) \\ let \ w = 50 t \\ u= \cos w \\ and, \ q = 4tu \\ I = \frac {dq} {dt} = \frac {dw} {dt} \times \frac {du} {dw} \times \frac {dq} {du} \\ w = 50 t, \ \frac {dw} {dt} = \frac {d} {dt} (50 t) = 50 \\ u = \cos w, \frac{du}{dw} = \frac {d} {dw} (\cos w) = -\sin w \\ q = 4tu, \frac {dq} {du} = \frac {d} {du} (4tu) = 4t \\ I = 0 - \frac {dq} {dt} = - \frac {dq} {dt} = - (50 \times -\sin w \times 4t) \ µC \\ I = 200t \sin 50 t \ µC \\ or \\ I = 200t \sin 50 t \times 10^{-6} \ C \\ I = 0.2t \sin 50 t \times 10^{-3} \ C \\ I = 0.2t \cos 50 t \ mC$