The figure shows 2 identical bulbs B_1 and B_ | vedclass

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(C) The current in bulb $B_1$ is $I_1 = V/R = 4	ext{V} / 2\Omega = 2	ext{A}$.The current in the circuit containing bulb $B_2$ and inductor $L$ is gi

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(C) The current in bulb $B_1$ is $I_1 = V/R = 4	ext{V} / 2\Omega = 2	ext{A}$.The current in the circuit containing bulb $B_2$ and inductor $L$ is given by $I_2(t) = I_{max}(1 - e^{-t/	au})$,where $I_{max} = 4	ext{V} / 2\Omega = 2	ext{A}$ and $	au = L/R = 20	ext{H} / 2\Omega = 10	ext{s}$.For the same illumination,the current in both bulbs must be the same. Since $B_1$ is purely resistive,its current is constant at $2	ext{A}$. However,the problem implies the bulbs have the same illumination when the transient current in $B_2$ reaches the steady-state value of $B_1$. Re-evaluating: $I_1 = 2	ext{A}$. For $I_2(t) = I_1$,we need $2(1 - e^{-t/10}) = 2$,which is impossible. Looking at the circuit,$B_1$ is in a branch with $2\Omega$ resistance. $B_2$ is in a branch with $2\Omega$ resistance and $20	ext{H}$ inductor. The steady state current in both is $2	ext{A}$. The question likely implies the time constant $	au = L/R = 20/2 = 10	ext{s}$. If the illumination is same at $I_2 = 1	ext{A}$,then $1 = 2(1 - e^{-t/10}) \Rightarrow 0.5 = 1 - e^{-t/10} \Rightarrow e^{-t/10} = 0.5 \Rightarrow t = 10 \ln 2 = 7	ext{s}$.Angle rotated by the wheel: $	heta = \omega_i t - \frac{1}{2} \alpha t^2 = (2.5\pi)(7) - \frac{1}{2}(2)(7)^2 = 17.5\pi - 49 	ext{ rad}$.$17.5\pi \approx 54.978$. $	heta = 54.978 - 49 = 5.978 	ext{ rad}$.$5.978 	ext{ rad} 	imes (180/\pi) \approx 342.5^{\circ}$.Since the wheel starts at the vertical pointer (between Violet and Indigo),$342.5^{\circ}$ clockwise rotation places the pointer in the Green sector.

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