Question

b only

v(t) Set up the differential equation and solve for the voltage in the circuit shown to the right for the following component values. a) i(t)-3x(t), b)i(t)-Su(t), R-200 Ω, L-4nH, i1(0)-0A R-1000 Ω, L-40 ml, iL(0)-2A

Answer 1

The given circuit will be easily solved in the Laplace domain.

In the Laplace domain, the circuit will be represented as

The given circuit will be easily solved in the Laplace domain in the Laplace domain the circuit will be represented as from the circuit we have u(t) = Ld_iL(t)/dt applying Laplace transform on the above equation v(s) = sLI_L (s) - i_L(0) equation according to the voltage - division principle L_L(s) = R/R + Sl I(s) from equation 1 and equation 2 v(s) = sL R/R + sL I (s) - i_L (0) rightarrow V(s) = s LR/R + sL I(s) - i_L(0) rightarrow (R + sL) v (s) = I (s) (R + sL) - i_L(0) (R + sL) applying inverse Laplace transforms rightarrow u(t) R + Ldv(t)/dt = R di(t)/dt + L D^2 I(T)/DT^2 - R_i L(0) delta (t) rightarrow v(t) + L/R dv(t)/dt = di(t)/dt + L D^2 I(T))/DT^2 - I_l (0) delta (T) i(t) = 3 u (t): R = 200 ohm L = 4 mH: i_L (0) = 0 A

From the circuit, we have

$v(t)=L\frac{\mathrm{d}i_L(t) }{\mathrm{d} t}$

Applying Laplace transform on the above equation,

$V(s)=sLI_L(s)-i_L(0)$ ....... Eq.1

According to the voltage-division principle,

$I_L(s)=\frac{R}{R+sL}I(s)$ ...... Eq.2

From Eq.1 and Eq.2,

$V(s)=sL\frac{R}{R+sL}I(s)-i_L(0)$

$\Rightarrow V(s)=\frac{sLR}{R+sL}I(s)-i_L(0)$ ........ Eq.3

$\Rightarrow (R+sL)V(s)=I(s)(R+sL)-i_L(0)(R+sL)$

Applying inverse Laplace transforms

$\Rightarrow v(t)R+L\frac{\mathrm{d} v(t)}{\mathrm{d} t}=R\frac{\mathrm{d} i(t)}{\mathrm{d} t}+L\frac{\mathrm{d}^2i(t)) }{\mathrm{d} t^2}-Ri_L(0)\delta(t)$

$\Rightarrow v(t)+\frac{L}{R}\frac{\mathrm{d} v(t)}{\mathrm{d} t}=\frac{\mathrm{d} i(t)}{\mathrm{d} t}+\frac{L}{R}\frac{\mathrm{d}^2i(t)) }{\mathrm{d} t^2}-i_L(0)\delta(t)$

(a) $i(t)=3u(t);\ R=200\Omega;\ L=4mH;\ i_L(0)=0A$

$I(s)=\frac{3}{s}$

From Eq.3

$\Rightarrow V(s)=\frac{s4\times10^{-3}\times200}{200+s4\times10^{-3}}\frac{3}{s}-0$

$\Rightarrow V(s)=\frac{0.8s}{(200+0.004s)}\frac{3}{s}$

$\Rightarrow V(s)=\frac{2.4}{(200+0.004s)}$

$\Rightarrow V(s)=\frac{600}{(s+50000)}$

$L^{-1}\left[\frac{1}{s+a} \right ]=e^{-at}$

Applying inverse Laplace transform

$v(t)=600e^{-50000t}V$

(b) $i(t)=5u(t);\ R=1000\Omega;\ L=40mH;\ i_L(0)=2A$

$I(s)=\frac{5}{s}$

From Eq.3

$\Rightarrow V(s)=\frac{s\times40\times10^{-3}\times1000}{1000+s40\times10^{-3}}\frac{5}{s}-2$

$\Rightarrow V(s)=\frac{200}{1000+0.04s}-2$

$\Rightarrow V(s)=\frac{5000}{25000+s}-2$

Applying inverse Laplace transforms

$v(t)=5000e^{-25000t}-2\delta(t)$