Study the circuits (a) and (b) shown in Fig. and answer the following questions. 1. Under which conditions would the rms currents in the two circuits be the same? 2. Can the rms current in circuit (b) be larger than that in (a)?

Key concept: Series RLC - Circuit 1. Equation of current: i=i_0 ( ); where i_0= V_0 Z 2. Equation of vpltage: From phasor diagram V= V^2_R+(V_L-V_C)^2 3. Impedance of the circuit: Z= R^2(X_L-X_C)^2 = R^2+ ( -1 )^2 4. Phase difference: From phasor diagram = V_L-V_C V_R = X_L-X_L R = -1 R = 2 -1 2 R 5. If net reactance is inductive: Circuit behaves as LR circuit 6. If net reactance is capacitive: Circuit begaves as CR circuit 7. If net reactance is zero: Means X = XL - XC = 0 ⇒ XL = XC, This is the condition of resonance. 8. At resononce (series resonant circuit), 1. XL = XC ⇒ Zmin = R, i.e., circuit behaves as a resistive circuit. 2. VL = VC ⇒ V = VR, i.e., whole applied voltage appeared across the resistance. 3. Phase difference: =0^ p.f.= =1 4. Power consumption: P=V_rmsi_rms=1 2 V_0i_0 5. Current in the circuit is maximum and it is i_0= V_0 R Let us first assume, rms current in circuit A = (Irms) A And rms current in circuit B = (Irms) B (I_rms)A= E_rms Z = E_rms R (I_rms)B= E_rms Z = E_rms R^2+(X_L-X_C)^2 1. When (Irms)A = (Irms)B R= R^2+(X_L-X_C)^2 ⇒ XL = XC, resonance condition If Erms in the two circuit are same, then at resonance the rms currnet in LCR will be same as that in R circuit (circuit A). 2. As Z (I_rms)_B (I_rms)_A = E_rms Z E_rms R = R R^2 (X_L-X_C)^2 = R Z 1 (I_rms)a (I_rms)b No, R . So, rms current in circuit (b) cannot be larger than that in (a).

Study the circuits (a) and (b) shown in Fig. and answer the follo…

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Question

Study the circuits (a) and (b) shown in Fig. and answer the following questions.

Under which conditions would the rms currents in the two circuits be the same?
Can the rms current in circuit (b) be larger than that in (a)?

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Answer

Key concept: Series RLC - Circuit

Equation of current: $\text{i}=\text{i}_0\sin(\omega\text{t}\pm\phi);\text{ where i}_0=\frac{\text{V}_0}{\text{Z}}$
Equation of vpltage: From phasor diagram $\text{V}=\sqrt{\text{V}^2_\text{R}+(\text{V}_\text{L}-\text{V}_\text{C})^2}$
Impedance of the circuit: $\text{Z}=\sqrt{\text{R}^2(\text{X}_\text{L}-\text{X}_\text{C})^2}=\sqrt{\text{R}^2+\Big(\omega\text{L}-\frac{1}{\omega\text{C}}\Big)^2}$
Phase difference: From phasor diagram $\tan\phi=\frac{\text{V}_\text{L}-\text{V}_\text{C}}{\text{V}_\text{R}}=\frac{\text{X}_\text{L}-\text{X}_\text{L}}{\text{R}}=\frac{\omega\text{L}-\frac{1}{\omega\text{C}}}{\text{R}}=\frac{2\pi\text{vL}-\frac{1}{2\pi\text{vC}}}{\text{R}}$
If net reactance is inductive: Circuit behaves as LR circuit
If net reactance is capacitive: Circuit begaves as CR circuit
If net reactance is zero: Means X = X_L - X_C = 0 ⇒ X_L = X_C, This is the condition of resonance.
At resononce (series resonant circuit),

X_L = X_C ⇒ Z_min = R, i.e., circuit behaves as a resistive circuit.
V_L = V_C ⇒ V = V_R, i.e., whole applied voltage appeared across the resistance.
Phase difference: $\phi=0^\circ\Rightarrow\ \text{p.f.}=\cos\phi=1$
Power consumption: $\text{P}=\text{V}_\text{rms}\text{i}_\text{rms}=\frac{1}{2}\text{V}_0\text{i}_0$
Current in the circuit is maximum and it is $\text{i}_0=\frac{\text{V}_0}{\text{R}}$

Let us first assume, rms current in circuit A = (I_rms) A

And rms current in circuit B = (I_rms) B

$(\text{I}_\text{rms})\text{A}=\frac{\text{E}_\text{rms}}{\text{Z}}=\frac{\text{E}_\text{rms}}{\text{R}}$

$(\text{I}_\text{rms})\text{B}=\frac{\text{E}_\text{rms}}{\text{Z}}=\frac{\text{E}_\text{rms}}{\sqrt{\text{R}^2+(\text{X}_\text{L}-\text{X}_\text{C})^2}}$

When (I_rms)A = (I_rms)B

$\text{R}=\sqrt{\text{R}^2+(\text{X}_\text{L}-\text{X}_\text{C})^2}$

⇒ X_L = X_C, resonance condition

If Erms in the two circuit are same, then at resonance the rms currnet in LCR will be same as that in R circuit (circuit A).

As $\text{Z}\geq\text{R}$

$\Rightarrow\ \frac{(\text{I}_\text{rms})_\text{B}}{(\text{I}_\text{rms})_\text{A}}=\frac{\frac{\text{E}_\text{rms}}{\text{Z}}}{\frac{\text{E}_\text{rms}}{\text{R}}}=\frac{\text{R}}{\sqrt{\text{R}^2}(\text{X}_\text{L}-\text{X}_\text{C})^2}=\frac{\text{R}}{\text{Z}}\leq1$

$\Rightarrow\ (\text{I}_\text{rms})\text{a}\geq(\text{I}_\text{rms})\text{b}$

No, $\text{R}\leq\text{Z}$. So, rms current in circuit (b) cannot be larger than that in (a).