- #1

User1265

- 29

- 1

- Homework Statement
- A signal generator, resistance 300Ω , produces a square-wave signal of frequency

1.0 MHz. The output of the generator is connected to an oscilloscope, with negligible

capacitance, by a coaxial cable of capacitance 1.5 × 10^-9F and negligible resistance.

Sketch the trace observed on the oscilloscope screen.

Give any necessary calculations.

- Relevant Equations
- time constant of Capacitor = RC

1/f = T

I started by working out the time constant of the Capacitor = RC = 300 * 1.5*10^-9 = 0.45 *10^-6

and the time period (t) of the square wave signal produced by the signal generator : 1/(1*10^6 )= 10^-6

Then I drawed the graph as follows:

https://www.physicsforums.com/data/attachments/234/234320-391937ca3e0804dfc6126d030d757447.jpg I sketched what I thought the circuit was for this Q below the graph. I unsure how the graph of the capacitor would look like for when the voltage applied by the singal generator is negative.

T /2 is roughly equal to one time constant so on the graph, at T/2 I showed the Chargge/Voltage/current of the capacitance at 63% roughly of the p.d applied across the capactior and the resistor.

However the worked solutions below show a different graph, along with some of the reasoning.

https://www.physicsforums.com/data/attachments/234/234321-6a47ec112bdd2f8099738f570ac71db3.jpg

1) I'm really confused on the reasoning given for the graph for the capacitor shown in the solutions, and thus I can't understand the graph. If it could be explained more clearly I would be grateful.

Also, Q1) Why does the graph of the capactior start at a negative value at t=0 , surely it would have no charge on its plates intially?

12) Am I correct that the capacitor, when the voltage is reversed by the signal generator, after T/2, the capactior reaches the same magntiude of charge (on its other plate this time) , but in a smaller time than the time constant - as in the reverse direction, the current reaching the capactior plate would be greater due to not having to go through the resistor.

and the time period (t) of the square wave signal produced by the signal generator : 1/(1*10^6 )= 10^-6

Then I drawed the graph as follows:

https://www.physicsforums.com/data/attachments/234/234320-391937ca3e0804dfc6126d030d757447.jpg I sketched what I thought the circuit was for this Q below the graph. I unsure how the graph of the capacitor would look like for when the voltage applied by the singal generator is negative.

T /2 is roughly equal to one time constant so on the graph, at T/2 I showed the Chargge/Voltage/current of the capacitance at 63% roughly of the p.d applied across the capactior and the resistor.

However the worked solutions below show a different graph, along with some of the reasoning.

https://www.physicsforums.com/data/attachments/234/234321-6a47ec112bdd2f8099738f570ac71db3.jpg

1) I'm really confused on the reasoning given for the graph for the capacitor shown in the solutions, and thus I can't understand the graph. If it could be explained more clearly I would be grateful.

Also, Q1) Why does the graph of the capactior start at a negative value at t=0 , surely it would have no charge on its plates intially?

12) Am I correct that the capacitor, when the voltage is reversed by the signal generator, after T/2, the capactior reaches the same magntiude of charge (on its other plate this time) , but in a smaller time than the time constant - as in the reverse direction, the current reaching the capactior plate would be greater due to not having to go through the resistor.