An inductor of inductance 10 text{ mH} having | vedclass

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(A) The voltage across an inductor in an $LR$ circuit is given by the equation $V_L = \varepsilon - iR$, where $\varepsilon$ is the battery $E$.$M$.$F

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(A) The voltage across an inductor in an $LR$ circuit is given by the equation $V_L = \varepsilon - iR$, where $\varepsilon$ is the battery $E$.$M$.$F$.,$i$ is the instantaneous current, and $R$ is the resistance of the inductor.Given: $\varepsilon = 1.0 	ext{ V}$, $R = 100 \ \Omega$.For current $i_1 = 2 	ext{ mA} = 2 	imes 10^{-3} 	ext{ A}$, the voltage across the inductor is:$V_{L1} = 1.0 - (2 	imes 10^{-3} 	imes 100) = 1.0 - 0.2 = 0.8 	ext{ V}$.For current $i_2 = 4 	ext{ mA} = 4 	imes 10^{-3} 	ext{ A}$, the voltage across the inductor is:$V_{L2} = 1.0 - (4 	imes 10^{-3} 	imes 100) = 1.0 - 0.4 = 0.6 	ext{ V}$.The ratio of the instantaneous voltages is:$\frac{V_{L1}}{V_{L2}} = \frac{0.8}{0.6} = \frac{4}{3}$.

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