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An LCR series circuit behaves like a dam...

An LCR series circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant ‘b’, the correct equivalence would be (Take, i = current, V = voltage, x = displacement, q = charge, F = force, v = velocity)

A

`qharrx,vltharrgti,FharrV`

B

`qharrv,iharrx,FharrV`

C

`qharrx,vharrV,Ftoi`

D

`qharrF,iharrx,Vtov`

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The correct Answer is:
To solve the problem of comparing an LCR series circuit with a damped harmonic oscillator, we will derive the equations for both systems and establish the equivalences. ### Step-by-Step Solution: 1. **Write the Equation for a Damped Harmonic Oscillator:** The equation for a damped harmonic oscillator (spring-mass system) is given by: \[ m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0 \] Here, \(m\) is the mass, \(b\) is the damping constant, \(k\) is the spring constant, \(x\) is the displacement, and \(\frac{dx}{dt}\) is the velocity. **Hint:** Identify the terms in the equation: mass, damping, and restoring force. 2. **Write the Equation for an LCR Series Circuit:** For an LCR circuit, applying Kirchhoff's voltage law gives: \[ L \frac{di}{dt} + Ri + \frac{q}{C} = 0 \] where \(L\) is the inductance, \(R\) is the resistance, \(C\) is the capacitance, \(i\) is the current, and \(q\) is the charge on the capacitor. **Hint:** Recognize that the terms represent inductive, resistive, and capacitive effects. 3. **Relate Current and Charge:** The current \(i\) is the rate of change of charge \(q\): \[ i = \frac{dq}{dt} \] Substitute this into the LCR equation: \[ L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = 0 \] **Hint:** Use the relationship between current and charge to express the equation in terms of charge. 4. **Compare the Two Equations:** Now we have two equations: - Damped oscillator: \(m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0\) - LCR circuit: \(L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = 0\) By comparing coefficients, we can establish the following equivalences: - \(m \leftrightarrow L\) (mass corresponds to inductance) - \(b \leftrightarrow R\) (damping constant corresponds to resistance) - \(k \leftrightarrow \frac{1}{C}\) (spring constant corresponds to the reciprocal of capacitance) **Hint:** Look for the corresponding terms in both equations to establish equivalences. 5. **Establish the Final Equivalences:** From the comparisons, we can summarize: - Charge \(q\) is analogous to displacement \(x\) - Current \(i\) is analogous to velocity \(\frac{dx}{dt}\) - Voltage \(V\) is analogous to force \(F\) **Hint:** Identify how each physical quantity in the LCR circuit corresponds to the quantities in the damped oscillator. ### Conclusion: The correct equivalences based on the analysis are: - \(q \leftrightarrow x\) - \(i \leftrightarrow \frac{dx}{dt}\) - \(V \leftrightarrow F\) - \(b \leftrightarrow R\) - \(m \leftrightarrow L\) Thus, the correct answer is the first option that matches these equivalences.

To solve the problem of comparing an LCR series circuit with a damped harmonic oscillator, we will derive the equations for both systems and establish the equivalences. ### Step-by-Step Solution: 1. **Write the Equation for a Damped Harmonic Oscillator:** The equation for a damped harmonic oscillator (spring-mass system) is given by: \[ m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0 ...
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Figure 6.36 shows a potentiometer with a cell of emf 2.0 V and internal resistance 0.4 Omega maintaining a potential drop across the resistor wire AB . A standard cell that maintains a constant emf of 1.02 V (for very moderate current up to emf mu A ) gives a balance point at 67.3 cm length of the wire. To ensure very low current is drawn the standard cell, a very high resistance of 600 k Omega is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf epsilon and the balance point found, similary, turns out to be at 82.3 cm length of the wire. a. What is the value of epsilon ? b.What purpose does the high resistance of 600 k Omega have? c. Is the balance point affected by this high resistance? d. Is the balance point affected by internal resistance of the driver cell? e. Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V ? f. Would the circuit work well for determining an extermely small emf, say of the order of a few mV (such as the typical emf of a thermocouple)? If not, how will you modify the circuit ?

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