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

inlays85k5 2021-11-23 Answered
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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Expert Answer

Huses1969
Answered 2021-11-24 Author has 7160 answers
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
Answered 2021-11-25 Author has 8448 answers

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:
\(\displaystyle{v}{\left({t}\right)}=-{5}{e}^{{-{500}{t}}}+{8}{e}^{{-{2000}{t}}}\)
\(\displaystyle{v}{\left({t}\right)}={0}\Rightarrow\)
\(\displaystyle{5}{e}^{{-{500}{t}}}={8}{e}^{{-{2000}{t}}}\)
\(\displaystyle{\frac{{{8}}}{{{5}}}}={\frac{{{e}^{{-{500}{t}}}}}{{{e}^{{-{2000}{t}}}}}}\)
\(\ln(1.6)=\ln(e^{1500t})\)
\(\displaystyle{t}={\frac{{{\ln{{\left({1.6}\right)}}}}}{{{1500}}}}\)
\(\displaystyle{t}={313.33}\mu{s}\)

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