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(A) Given: Resistance $R = 50 \Omega$,$EMF$ $\varepsilon = 5 V$,current $i = 50 mA = 50 	imes 10^{-3} A$,and time $t = 0.1 s$.The growth of current i

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$\frac{5}{\ln(2)}$

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(A) Given: Resistance $R = 50 \Omega$,$EMF$ $\varepsilon = 5 V$,current $i = 50 mA = 50 	imes 10^{-3} A$,and time $t = 0.1 s$.The growth of current in an $LR$ circuit is given by the formula: $i = i_0(1 - e^{-\frac{Rt}{L}})$,where $i_0 = \frac{\varepsilon}{R}$ is the steady-state current.First,calculate the steady-state current: $i_0 = \frac{5 V}{50 \Omega} = 0.1 A = 100 mA$.Substitute the values into the growth equation: $50 	imes 10^{-3} = 100 	imes 10^{-3} (1 - e^{-\frac{50 	imes 0.1}{L}})$.Simplify: $0.5 = 1 - e^{-\frac{5}{L}}$.Rearrange: $e^{-\frac{5}{L}} = 1 - 0.5 = 0.5$.Taking the natural logarithm on both sides: $-\frac{5}{L} = \ln(0.5) = \ln(2^{-1}) = -\ln(2)$.Therefore,$L = \frac{5}{\ln(2)} H$.

$A$ coil of resistance $50 \Omega$ is connected across a $5.0 V$ battery. If the current in the coil is found to be $50 mA$ after the time $t=0.1 s$ of the battery being connected,then the inductance of the coil is:

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