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The switch in Figure is closed on a at time t = 0.
What is the ratio $$\mathscr{E}_L/\mathscr{E}$$ of the inductor’s self-induced emf to the battery’s emf (a) just after t = 0 and
(b) at $$t = 2.00\tau
_L$$? (c) At what multiple of $$\tau
_L$$ will $$\mathscr{E}_L/\mathscr{E}=0.500$$?
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Solution
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(a) Immediately after the switch is closed, $$\varepsilon-\varepsilon_{L}=i R .$$
But $$i=0$$ at this instant, so $$\varepsilon_{L}=\varepsilon,$$ or $$\varepsilon_{L}
/ \varepsilon=1.00$$
(b) $$\varepsilon_{L}(t)=\varepsilon
e^{-t / \tau_{L}}=\varepsilon e^{-2.0 \tau_{L} / \tau_{L}}=\varepsilon
e^{-2.0}=0.135 \varepsilon,$$ or $$\varepsilon_{L} / \varepsilon=0.135$$
(c) From $$\varepsilon_{L}(t)=\varepsilon
e^{-t / \tau_{L}}$$ we obtain
$$\dfrac{t}{\tau_{L}}=\ln
\left(\dfrac{\varepsilon}{\varepsilon_{L}}\right)=\ln 2 \Rightarrow t=\tau_{L}
\ln 2=0.693 \tau_{L} \Rightarrow t / \tau_{L}=0.693$$
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Q1
The switch in Figure is closed on a at time t = 0.
What is the ratio $$\mathscr{E}_L/\mathscr{E}$$ of the inductor’s self-induced emf to the battery’s emf (a) just after t = 0 and
(b) at $$t = 2.00\tau
_L$$? (c) At what multiple of $$\tau
_L$$ will $$\mathscr{E}_L/\mathscr{E}=0.500$$?
What is the ratio $$\mathscr{E}_L/\mathscr{E}$$ of the inductor’s self-induced emf to the battery’s emf (a) just after t = 0 and
(b) at $$t = 2.00\tau
_L$$? (c) At what multiple of $$\tau
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Q4
The switches in figure (a) and (b) are closed at t = 0 and reopened after a long time at t = t0.
(a) The charge on C just after t = 0 is εC.
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(b) The charge on C long after t = 0 is εC.
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(d) The current in L long after t = t0 is ε/R.