# The current in a 50 mH inducator is known to be i=120mA,t\le 0;

The current in a 50 mH inducator is known to be
$$\displaystyle{i}={120}{m}{A},{t}\le{0};$$ $$\displaystyle{i}={A}{1}{e}^{{-{500}{t}}}+{A}{2}{e}^{{-{2000}{t}}}{A},{t}\geq{0}$$.
The voltage across the inducator (passive sign convention) is 3 V at $$\displaystyle{t}={0}$$.
a) Find the expression for the voltage across the inducator for $$\displaystyle{t}{>}{0}$$.
b) Find the time, greater then zero, when the power at the terminals of the inductor is zero.

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Huses1969
a) We know that $$\displaystyle{i}{\left({0}\right)}={120}{m}{A}$$ so we can write
$$\displaystyle{i}{\left({0}\right)}={A}_{{{1}}}{e}^{{{0}}}+{A}_{{{2}}}{e}^{{{0}}}={A}_{{{1}}}+{A}_{{{2}}}={0.12}{A}$$
We also know that $$\displaystyle{v}{\left({0}\right)}={3}{V}$$ so we write
$$\displaystyle{\frac{{{d}{i}}}{{{\left.{d}{t}\right.}}}}=-{500}{A}_{{{1}}}{e}^{{-{500}{t}}}-{2000}{A}_{{{2}}}{e}^{{-{2000}{t}}}$$
$$\displaystyle{v}={L}{\frac{{{d}{i}}}{{{\left.{d}{t}\right.}}}}=-{25}{A}_{{{1}}}{e}^{{-{500}{t}}}-{100}{A}_{{{2}}}{e}^{{-{2000}{t}}}$$
$$\displaystyle{v}{\left({0}\right)}=-{25}{A}_{{{1}}}{e}^{{{0}}}-{100}{A}_{{{2}}}{e}^{{{0}}}=-{25}{A}_{{{1}}}-{100}{A}_{{{2}}}={3}{V}$$
Now we have two equations with two unknowns:
$$\displaystyle{A}_{{{1}}}+{A}_{{{2}}}={0.12}$$
$$\displaystyle-{25}{A}_{{{1}}}-{100}{A}_{{{2}}}={3}$$
We solve this and get
$$\displaystyle{A}_{{{1}}}={0.2}$$ and $$\displaystyle{A}_{{{2}}}=-{0.08}$$
Now we can write the expression for the voltage
$$\displaystyle{v}{\left({t}\right)}=-{25}{\left({0.2}\right)}{e}^{{-{500}{t}}}-{100}{\left(-{0.08}\right)}{e}^{{-{2000}{t}}}$$
$$\displaystyle{v}{\left({t}\right)}=-{5}{e}^{{-{500}{t}}}+{8}{e}^{{-{2000}{t}}}{V}$$
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Coon2000

b)
$$\displaystyle{p}={v}{i}$$ so we search for when the voltage current are zero
$$\displaystyle{i}{\left({t}\right)}={0.2}{e}^{{-{500}{t}}}-{0.08}{e}^{{-{2000}{t}}}$$
$$\displaystyle{i}{\left({t}\right)}={0}\Rightarrow$$
$$\displaystyle{0.2}{e}^{{-{500}{t}}}={0.08}{e}^{{-{2000}{t}}}$$
$$\displaystyle{\frac{{{0.2}}}{{{0.08}}}}={\frac{{{e}^{{-{2000}{t}}}}}{{{e}^{{-{500}{t}}}}}}$$
$$\ln(2.5)=\ln(e^{-1500t})$$
$$\displaystyle{t}={\frac{{-{\ln{{\left({2.5}\right)}}}}}{{{1500}}}}$$
$$\displaystyle{t}=-{6.1}\cdot{10}^{{-{4}}}{s}$$
Whitch is not greater then zero. Let`s look at the voltage now:
$$\displaystyle{v}{\left({t}\right)}=-{5}{e}^{{-{500}{t}}}+{8}{e}^{{-{2000}{t}}}$$
$$\displaystyle{v}{\left({t}\right)}={0}\Rightarrow$$
$$\displaystyle{5}{e}^{{-{500}{t}}}={8}{e}^{{-{2000}{t}}}$$
$$\displaystyle{\frac{{{8}}}{{{5}}}}={\frac{{{e}^{{-{500}{t}}}}}{{{e}^{{-{2000}{t}}}}}}$$
$$\ln(1.6)=\ln(e^{1500t})$$
$$\displaystyle{t}={\frac{{{\ln{{\left({1.6}\right)}}}}}{{{1500}}}}$$
$$\displaystyle{t}={313.33}\mu{s}$$