Given condition

The phase difference between alternating current and emf is $\phi = \frac{\pi}{2}.$

This means current and voltage are exactly $90^\circ$ out of phase.

Recall phase relations for basic AC circuits

• Pure resistor $R$: current and voltage are in phase $\phi = 0.$

• Pure inductor $L$: current lags voltage by $\phi = \frac{\pi}{2}.$

• Pure capacitor $C$: current leads voltage by $\phi = \frac{\pi}{2}.$

So, for a pure inductor or pure capacitor, the magnitude of phase difference is indeed $\frac{\pi}{2}$.

Check each option

Option B: $C$ alone

For a pure capacitor, $\phi = \frac{\pi}{2}$ (current leads emf by $90^\circ$).

So this can be the constituent.

Option C: $L$ alone

For a pure inductor, $\phi = \frac{\pi}{2}$ (current lags emf by $90^\circ$).

So this can be the constituent.

Option D: $L, C$

In an ideal $LC$ circuit, net reactance is $X = X_L - X_C.$ Since there is no resistance, impedance is purely reactive: $Z = j(X_L - X_C).$ Hence current and emf remain $90^\circ$ out of phase (except at resonance where impedance becomes zero and the usual phase relation becomes trivial).

So this can be a constituent combination giving phase difference $\frac{\pi}{2}$.

Option A: $R, L$

For an $RL$ circuit, $\tan \phi = \frac{X_L}{R}.$ Therefore, $\phi < \frac{\pi}{2}$ for any finite resistance $R>0$.

To get exactly $\phi = \frac{\pi}{2},$ we would need $R=0,$ but then it is no longer an $RL$ circuit with resistance present.

Hence an $RL$ circuit cannot have phase difference exactly $\frac{\pi}{2}$.

Conclusion

The constituent that cannot be present is: $\boxed{A}$