I'm interested in the voltage, In the picture I attached there is a supply voltage of 15V and for a given question let's say the RA needs 6V to work, when the current through the photodiode is sufficient to give 6V to the RA does this mean that the voltage drop across the diode is 15-6 = 9 ?
Hi all,
I'm having a little bit of trouble understanding photodiodes in reverse bias. Say for instance I have the circuit shown, If I'm asked for a certain voltage required to switch on RA I understand that this corresponds to a certain current being allowed through the diode which corresponds...
ok, this helps me understand power dissipated but in terms of the voltage drop, is the only way to explain it that the voltage on one side of the parallel component is x and on the other side is y so the drop across each resistor is x-y ? I can't conceptually understand how the voltage drops...
Hi all,
I think I know the answer to this question but I'm having trouble explaining why it is so.
If I have a circuit with a fixed resistor connected in parallel with a reverse biased diode, I believe the voltage drop across each will be the same. Is this correct? If so can someone explain the...
Yes, this all makes sense conceptually (really well explained) and I am more than happy that the maths agrees nicely, my question is what is it that causes the normal force to be less than the weight force at the top of the hill? I understand that if the car is traveling slowly more normal force...
Yep that makes sense to me, the car that has fallen from a height is exerting a greater force on the road and therefore the force the road exerts back is higher. But in the case of the car on the hill are you suggesting that the car's acceleration in the vertical direction is decreasing? and...
Ok, can you potentially explain that further? when the car is exactly on top of the hill, if it were stationery then the normal reaction force would be equal to the weight force, yet when moving it is less than the weight force. Doesn't all the weight force go into accelerating the car down?
Hi, I have been studying up on circular motion in the vertical plane and I am happy with all the math/theory that I have read explaining the normal force and weight force and the feeling of being "heavier" and "lighter". My question is about the top of the hill/ride, The forces acting on you at...
Ok cool that sounds reasonable, so am I correct in saying that the component of the normal caused by gravity is exactly canceled by the weight force, and the reason that there is a horizontal component of the normal providing centripetal acceleration is due to the normal reaction of the banked...
Ok, I think this is getting me close to an understanding, It's just hard to get my head around where that extra reaction force comes from? If there is no friction, it can't be friction from the road, and from what i can understand, it can't be due to a reaction from the weight force because the...
I think this is where the confusion lies, you say the normal force has to have a horizontal component to match/provide the centripetal acceleration but I think I'm having trouble with the cause/effect here? I'm thinking from the point of view that the horizontal component of the normal is caused...