Question

Hi, I had an experiment about electrical and mechanical energy.

Charging difference capacitance of capacitor and when it connect the motor, the weights move up. So, I need a efficency of each trial.

And, I have two equations from the class and prediction graph. But, I don't know how to find prediction equation with physical concepts.

Could you explain about that?

Thank you very much.

Answer 1

So here a capacitor is being charged and then used to drive a motor which lifts a weight up. In the experiment, the efficiency of the motor is being measured, i.e. the efficiency with which, it converts the electrical energy supplied to it to mechanical work/energy in lifting the weight.

When a potential difference of V is reached across a capacitor while getting charged, the charge on each plate of the capacitor is Q = C*V, where C is the capacitance.

During the charging, if v be the potential difference across the plates and q be the charge at any instant, the work done in increasing the charge by dq is given by

$dW = v\times dq$

So the net work done in charging the capacitor to Q which is the electrical energy stored in the capacitor is given by,

$W = \int_{0}^{Q} v\times dq = \int_{0}^{Q} \frac{q}{C}\times dq = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}CV^2$

When the capacitor gets discharged through the motor, this energy is supplied to the motor.

The energy spent or the work done by the motor in lifting the mass m through a height x is the change in gravitational potential energy of the mass, which is mgx.

Since there will be losses, a fraction of the electrical energy available to the motor will be used to do this mechanical work. i.e.

$m\times g\times x = e \times \frac{1}{2}CV^2 \newline \newline \implies V_y = e\times V_E \newline \newline where \hspace{4pt}e\hspace{4pt} is\hspace{4pt} the\hspace{4pt} efficiency.$

So the efficiency will be the slope of the $V_y \hspace{4pt}vs \hspace{4pt} V_E$ curve that can be drawn by measuring the lift and the voltage across the capacitor in each experiment and charging the capacitor to higher voltages in each trial within the capacitor rating.