Ratch, your addressing questions to which I already know the answer. What I wanted to know was the voltage across the transistor when the LDR as at the resistances given, and why it is so. I now know that because the transistor is off 10V of potential is stored in the wire, and that 10V of...
So basically, given the tiny current that comes about as a result of the massive resistance of the transistor, and given the tiny voltage drop across Rc that comes with this current, I can essentially look at that path through Rc to Vout as an extension of the cell.
Yes. Given that potential is lost, however small an amount, across Rc, current would have to flow through it. This current cannot then flow through the transistor to lose its potential at earth..
(I now see what you're getting at. Okay, The current is actually very small.)
But still, there is...
I already have done part A. That was the simple part it's just solving a voltage div equation for Vout = 0.65V.
b) and c) were the problem, but I think I understand and my possible understanding lines up with the answers given. Are you able to tell me if what I said in my last post is correct?
So does current only flow through the left side of the circuit? As that is the only path along which current can flow in light of infinite resistance from the inactive transistor and the wire that goes to nowhere.
Edit. Potential brainwave. For partis and b do you just work out the voltage as...
Voltage division equation.
The Attempt at a Solution
From a) I know that Rx = 86k ohms.
Part b) and c) befuddle me. I think it is relevant to the transistor being on and off.
In b) I would think that because Rx...