Two inductors L_1 (inductance 1 text{ mH},int | vedclass

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(B) At $t=0$,the inductors act as open circuits because they oppose the change in current. Thus,current flows only through the $12 	ext{ } \Omega$ re

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$8$

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(B) At $t=0$,the inductors act as open circuits because they oppose the change in current. Thus,current flows only through the $12 	ext{ } \Omega$ resistor.$I_{\min} = \frac{V}{R} = \frac{5}{12} 	ext{ A}$.At $t ightarrow \infty$,the inductors act as ideal conducting wires (short circuits) because the current becomes steady. The circuit now consists of three resistors in parallel: $3 	ext{ } \Omega$,$4 	ext{ } \Omega$,and $12 	ext{ } \Omega$.The equivalent resistance $R_{	ext{eff}}$ is given by:$\frac{1}{R_{	ext{eff}}} = \frac{1}{3} + \frac{1}{4} + \frac{1}{12} = \frac{4+3+1}{12} = \frac{8}{12} = \frac{2}{3} 	ext{ } \Omega^{-1}$.So,$R_{	ext{eff}} = 1.5 	ext{ } \Omega$.The maximum current is $I_{\max} = \frac{V}{R_{	ext{eff}}} = \frac{5}{1.5} = \frac{10}{3} 	ext{ A}$.The ratio is $\frac{I_{\max}}{I_{\min}} = \frac{10/3}{5/12} = \frac{10}{3} 	imes \frac{12}{5} = 2 	imes 4 = 8$.

Column $I$	Column $II$
$(A)$ $A$ charged capacitor is connected to the ends of the wire	$(p)$ $A$ constant current flows through the wire
$(B)$ The wire is moved perpendicular to its length with a constant velocity in a uniform magnetic field perpendicular to the plane of motion	$(q)$ Thermal energy is generated in the wire
$(C)$ The wire is placed in a constant electric field that has a direction along the length of the wire	$(r)$ $A$ constant potential difference develops between the ends of the wire
$(D)$ $A$ battery of constant emf is connected to the ends of the wire	$(s)$ Charges of constant magnitude appear at the ends of the wire

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