Let the two figures be the standard AC circuits:

• (a): a series $RLC$ circuit connected to an AC source
• (b): a parallel $RLC$ circuit connected to an AC source

We compare their rms currents drawn from the source.

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1. Current in series $RLC$ circuit (a)

For a series $RLC$ circuit, the impedance is

$$Z_s = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$$

Hence the rms current is

$$I_a = \frac{V}{Z_s}$$

Since

$$Z_s \ge R,$$

we get

$$I_a \le \frac{V}{R}$$

with equality at resonance:

$$\omega L = \frac{1}{\omega C}$$

So at resonance,

$$I_a = \frac{V}{R}$$

which is maximum for the series circuit.

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2. Current in parallel $RLC$ circuit (b)

For a parallel $RLC$ circuit, source current is found from admittance:

$$Y_p = \sqrt{\left(\frac{1}{R}\right)^2 + \left(\omega C - \frac{1}{\omega L}\right)^2}$$

Thus rms current drawn from source is

$$I_b = V Y_p = V\sqrt{\left(\frac{1}{R}\right)^2 + \left(\omega C - \frac{1}{\omega L}\right)^2}$$

Since

$$Y_p \ge \frac{1}{R},$$

we get

$$I_b \ge \frac{V}{R}$$

with equality at resonance.

At resonance,

$$\omega C = \frac{1}{\omega L}$$

so

$$I_b = \frac{V}{R}$$

which is minimum for the parallel circuit.

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3. Comparison of $I_a$ and $I_b$

From above,

$$I_a \le \frac{V}{R}, \qquad I_b \ge \frac{V}{R}$$

Therefore,

$$I_b \ge I_a$$

for all frequencies, and equality holds only at resonance.

So:

• circuit (b) can be larger than circuit (a) away from resonance
• they are not always equal
• circuit (b) can never be larger is false
• at resonance, they are equal, not less

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4. Check each option

Option A

The rms current in circuit (b) can be larger than that in (a)

This is true because away from resonance,

$$I_b > I_a$$

Option B

The rms current in figure (a) is always equal to that in figure (b)

This is false. They are equal only at resonance.

Option C

The rms current in circuit (b) can never be larger than that in (a)

This is false.

Option D

At resonance, current in (b) is less than that in (a)

This is false because at resonance,

$$I_a = I_b = \frac{V}{R}$$

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5. Final answer

The correct option is:

$$\boxed{A}$$

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6. Comparison with stored answer

Stored correct answer: $C$

Our derived answer is $A$, so the stored answer does not match the physics of standard series and parallel $RLC$ circuits. In fact, the comparison shows:

$$I_b \ge I_a$$

not the reverse.