About critical damping resistance of a ballistic galvanometer

  • #1
phymath7
48
4
Homework Statement
In my physics lab experiment,I need to find out the critical damping resistance of a ballistic galvanometer (The associated circuit is given in the attempt section) by drawing ##\lambda## vs. ##R_2## graph where ##\lambda## is the logarithmic decrement of deflection of galvanometer .
Relevant Equations
$$\omega=\sqrt{\omega_0{}^2 -\frac{\gamma^2}{4}}$$
where ##\omega## is the damped angular frequency and ##\omega_0## is the undamped angular frequency
The differential equation of the motion of the galvanometer(wrt time):
$$\ddot \theta+\gamma\dot \theta +k^2\theta=0$$
Relation between ##\lambda## and ##\gamma ## is:
$$\lambda=\frac{\gamma T}{4}$$
Where
$$\gamma =\frac{\beta +\frac {a} {R_2 +G}}{I}$$
##\beta## and 'a' are constant,G is the galvanometer resistance and T is the time period.
At critical condition, ##\omega=0## so time period will be infinite and so will be ##\lambda##.Therefore, the critical resistance will be the corresponding resistance(plus galvanometer resistance)of the asymptote of ##\lambda## vs. ##R_2## graph(the graph is a rectangular hyperbola).
But here's where I'm stuck.How am I supposed to find the asymptote of the graph only having the observed data and not the explicit function?
20230826_165720.jpg
 
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  • #2
phymath7 said:
Where
$$\gamma =\frac{\beta +\frac {a} {R_2 +G}}{I}$$
What is I? At first I thought it was a typo for T but that doesn't seem to make sense.
 
  • #3
haruspex said:
What is I? At first I thought it was a typo for T but that doesn't seem to make sense.
The denominator "I" is the moment of inertia of the coil of galvanometer.
 
  • #4
phymath7 said:
The denominator "I" is the moment of inertia of the coil of galvanometer.
If you fit the ##(R_2,\lambda)## data to that equation, can you deduce ##\beta/I, a/I, G##?
 
  • #5
haruspex said:
If you fit the ##(R_2,\lambda)## data to that equation, can you deduce ##\beta/I, a/I, G##?
Did you write the explicit relation between ##\lambda## and ##R_2## ?Do you see how complex this one? I am afraid that it's beyond my (and even for those who are more advanced) capability to do the fitting.
 
  • #6
phymath7 said:
Did you write the explicit relation between ##\lambda## and ##R_2## ?Do you see how complex this one? I am afraid that it's beyond my (and even for those who are more advanced) capability to do the fitting.
Assuming you have some idea of the value of G, put that in a spreadsheet and set up columns of values of ##1/\gamma, 1/(R_2+G)##.
Have the spreadsheet produce a linear regression, showing the goodness of fit.
Tweak G to maximise the fit.
 
  • #7
haruspex said:
Assuming you have some idea of the value of G, put that in a spreadsheet and set up columns of values of ##1/\gamma, 1/(R_2+G)##.
Have the spreadsheet produce a linear regression, showing the goodness of fit.
Tweak G to maximise the fit.
I need the relationship between ##\lambda## and ##R_2## which represents a hyperbola.How am I supposed to get that?
 

Related to About critical damping resistance of a ballistic galvanometer

What is critical damping resistance in a ballistic galvanometer?

Critical damping resistance in a ballistic galvanometer is the specific resistance value at which the galvanometer returns to zero without oscillating after a deflection. It ensures the quickest return to equilibrium without overshooting, optimizing the measurement accuracy for transient currents.

How is critical damping resistance calculated for a ballistic galvanometer?

Critical damping resistance (Rc) is calculated using the formula Rc = 2√(L/C), where L is the inductance and C is the capacitance of the galvanometer circuit. This formula ensures that the system is critically damped, providing optimal response characteristics.

Why is critical damping important in a ballistic galvanometer?

Critical damping is important in a ballistic galvanometer because it ensures that the pointer returns to zero as quickly as possible without oscillating. This is crucial for accurate measurement of transient currents, as it minimizes the time needed for the galvanometer to stabilize and prevents errors caused by overshooting and oscillations.

What happens if the resistance is lower or higher than the critical damping resistance?

If the resistance is lower than the critical damping resistance, the galvanometer will be underdamped, leading to oscillations and longer settling times. If the resistance is higher, the system will be overdamped, causing the pointer to return to zero more slowly than necessary, which can delay measurements and reduce efficiency.

Can the critical damping resistance be adjusted in a ballistic galvanometer?

Yes, the critical damping resistance can be adjusted in a ballistic galvanometer by changing the resistance in the circuit. This can be done using variable resistors or by adding/removing fixed resistors to achieve the desired damping effect, ensuring optimal performance for specific measurement conditions.

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