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Q11Q

Expert-verifiedFound in: Page 935

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Question: Figure 31-25 shows the current i and driving emf**${{\mathit{\omega}}}_{{\mathbf{d}}}$

- Relative to the emf curve, the current curve will shift rightward and amplitude of current will increase if $L$ is increased.
- Relative to the emf curve, the current curve will shift rightward and amplitude of current will increase if $C$ is increased.
- Relative to the emf curve, the current curve will shift rightward and amplitude of current will increase if ${\omega}_{d}$ is increased.

The graph 31.25 of current and emf is shown.

**We use the equation of amplitude of current for the RLC circuit and the equation of phase shift to solve this problem. Using that equation, we can imagine the change in amplitude of the current or phase shift due to an increase in inductance and capacitance of driving angular frequency.**

Formulae:

The current flowing in the RLC circuit, ${\mathit{I}}{\mathbf{=}}{\mathbf{}}\frac{{\mathbf{E}}_{\mathbf{m}}}{\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\left({X}_{L}-{X}_{C}\right)}^{\mathbf{2}}}}$ (i)

The reactance of the inductor in the RLC circuit, ${{\mathit{X}}}_{{\mathbf{L}}}{\mathbf{=}}{{\mathit{\omega}}}_{{\mathbf{d}}}{\mathit{L}}$ (ii)

The reactance of the capacitance in the RLC circuit, ${{\mathit{X}}}_{{\mathbf{C}}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{1}}{{\mathbf{\omega}}_{\mathbf{d}}\mathbf{C}}$ (iii)

The tangent value of the phase angle of the RLC oscillations, ${\mathit{t}}{\mathit{a}}{\mathit{n}}{\mathit{\varphi}}{\mathbf{=}}{\mathbf{}}\frac{{\mathbf{X}}_{\mathbf{L}}\mathbf{-}{\mathbf{X}}_{\mathbf{C}}}{\mathbf{R}}$ (iv)

The angular or driving frequency of RLC circuit, ${{\mathit{\omega}}}_{{\mathbf{d}}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{1}}{\sqrt{\mathbf{L}\mathbf{C}}}$ (v)

From equation (ii), we can say that when we increase the value of inductance, the inductive reactance ${X}_{L}$ will increase.

So from equation (iv), when ${X}_{L}$ increases, it will increase the phase shift. Hence, the current graph will shift rightward.

Now, from equation (i) of current amplitude, we can say that when ${X}_{L}$ increases, the graph shifts towards right to attain the condition for resonance

Hence, $\left({X}_{L}-{X}_{C}\right)$ decreases as ${X}_{L}$ increases. This increases $L$ .

Therefore, the increase in $L$ will increase the amplitude of the current.

From equation (ii), we can say that when we increase the value of capacitance, the capacitive reactance ${X}_{C}$ will decrease.

So from equation (iv), when ${X}_{C}$ decreases, it will increase the phase shift. Hence, the current graph will shift rightward.

Now, from equation (i) of current amplitude, we can say that the increase in C will decrease the value of ${X}_{C}$. However, the graph is shifting towards the right; it means it is closer to attain the condition for equilibrium.

Therefore, there would be increase in the amplitude of current.

From equation (v), when we increase the value of driving frequency, the product will decrease.

So, the circuit will be closer to resonance. Hence, the current and emf will become more in the phase means the current graph will shift rightward.

Also, the amplitude of current will be near to maximum, so, when ${\omega}_{d}$ increases, amplitude of current will increase.

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