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Question
How is peak current related to peak EMF and external resistance?
Options
\[I_0=\varepsilon_0+R\]
\[I_0=\frac{\varepsilon_0}{R}\]
\[I_0=\varepsilon_0R\]
\[I_0=\frac{R}{\varepsilon_0}\]
MCQ
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Solution
Peak current is the current amplitude through resistance \[R\]. It equals peak EMF divided by resistance: \[I_0=\frac{\varepsilon_0}{R}\].
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