Two identical inductors are connected in two different configurations P and Q, where a time varying current \( I(t) \) is flowing, as shown in the figure. The induced emf between points a and b for configuration P is \( E_P \) and that for configuration Q is \( E_Q \). The ratio \( E_P/E_Q \) is : [Neglect the effect of mutual inductance.]

In configuration P, three identical inductors are in series, so \( L_P = 3L \). In configuration Q, three identical inductors are in parallel, so \( L_Q = L/3 \). The induced emf \( E = L \frac{dI}{dt} \), thus \( E_P/E_Q = L_P/L_Q = 9 \), wait recalculating: configuration P has 3 in series? Looking closely at figure P, there are 3 inductors in series, configuration Q has 3 in parallel. \( 3L / (L/3) = 9 \)... wait let's check options. Option 1 is 1/4.