At resonance, current in (B) is less than that in (A)
The r.m.s. current in (B) can be larger than that in (A)
The r.m.s. current in (A) is always equal to that in (B)
The r.m.s. current in (B) can never be larger than that in (A)
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The Correct Option isD
Solution and Explanation
Step 1: Understanding the Concept
Circuit A has only a 30 ohm resistor on a 210 V, 50 Hz supply. Circuit B has the same resistor in series with a 40 mH inductor and a 0.4 microfarad capacitor on the same supply.
Step 2: Current in A
\[ I_A=\frac{210}{30}=7\ \text{A} \]
Step 3: Impedance of B
\[ Z_B=\sqrt{R^2+(X_L-X_C)^2}\ \ge R=30\ \Omega \]
So
\[ I_B=\frac{210}{Z_B}\le\frac{210}{30}=7\ \text{A} \]
The reactances make \(Z_B\) larger than \(R\), or at best equal to it when \(X_L=X_C\).
Step 4: Numbers here
\(X_L=2\pi(50)(0.04)\approx12.6\ \Omega\) and \(X_C=\frac{1}{2\pi(50)(0.4\times10^{-6})}\approx7960\ \Omega\). So \(Z_B\) is large and \(I_B\) is very small.
Step 5: Check the options
At resonance the currents would be equal, not less, so (A) is wrong. Since \(I_B\le I_A\) always, (B) is wrong. (C) is wrong because they are equal only at resonance. So (D): the current in B can never exceed that in A.
Final Answer:
The impedance of B is never smaller than 30 ohm, so its current never exceeds the 7 A in A, option (D).
\[ \boxed{\text{(D)}} \]
