resistors
In the given circuit, the variable resistor R = 1MΩ is connected as shown to a load RL = 1MΩ. The
variable resistor is adjusted until the voltage V
across the load is 8 V when the input voltage Vin =
27 V. What is the position of the wiper arm in percentage?
This is the question. I have solve it and my answer is 42%, but the solution said 36%.
My Calculation
It's simply telling you that the circuit as drawn will never have the given current running through it.
Consider the following situation: Replace the inductor by a capacitor (capacitor discharge might be more familiar to you), and the current function by
$$ I(t) = t $$
Now calculate voltages - they won't add up. Why? Because the current function makes no sense. The complete discharging process of a capacitor through a resistor is completely defined by the capacity, resistance and the voltage across it at time \$t = 0\$.
Similarly, the 'discharging' of an inductor through a resistor is completely determined by the inductance, resistance and the current through it at time \$t = 0\$.
If you are now given a time-dependent current function, you are overspecifying the system. That function may be right, but (like in your question) it may be wrong for the given circuit, so you arrive at contradictory solutions.
Note that there is a way to keep the question/function/values like they are now and make it consistent again, by adding an ideal current source into the circuit which satisfies the given current function. The strange voltage that you couldn't explain is then simply found across this current source:
simulate this circuit – Schematic created using CircuitLab
Alternatively, simply assume that the current function actually is right at \$t=0\$, that is, \$I_0 = 50\$. The discharge current of an inductor through a resistor is $$I(t) = I_0e^{-\frac{R}{L}t}$$, so the correct current function for the circuit as shown in your question is $$I(t) = 50e^{-20000t}$$. Do your voltage calculations again - they will now work out.
From your description, I believe this is what you want to say (and be disciplined enough to tell the details properly).
This is on basic diode circuit analysis. For an ideal diode in this configuration, the load resistor has a zero voltage across it when the AC input is in the positive half cycle. Why is this the case? Your load is in parallel with a short.
What happens? You can look at it in two ways. First, electrical current takes the path of least resistance. It will only run through the short. No current means no voltage for an Ohmic resistor (remember Ohm's Law). And second, your load resistor is in a parallel configuration with a short. A voltage across a short is always zero volts. And in a parallel configuration, the voltage must be the same for each branch.
What is the consequence then for the negative half cycle?
Your diode is in a reverse bias configuration. It means your diode behaves like an open. There is a voltage drop across the load resistor because the current passes through the load resistor and not through the diode.
simulate this circuit – Schematic created using CircuitLab
Best Answer
I approached it looking at the circuit as the following with \$R=1\:\textrm{M}\Omega\$:
simulate this circuit – Schematic created using CircuitLab
The central node voltage of such a resistor arrangement is easy to memorize:
$$V_o=\frac{V_1\cdot R_2\cdot R_3 + V_2\cdot R_1\cdot R_3+V_3\cdot R_1\cdot R_2}{R_1\cdot R_2+R_1\cdot R_3+R_2\cdot R_3}$$
In this case, \$V_1=V_2=0\:\textrm{V}\$, so:
$$\begin{align*} V_O&=\frac{V_3\cdot D\cdot R\cdot R}{D\cdot R\cdot R+D\cdot R\cdot \left(1-D\right)\cdot R+R\cdot \left(1-D\right)\cdot R}\\\\ &=\frac{V_3\cdot D\cdot R^2}{\left(D+D\cdot \left(1-D\right)+\left(1-D\right)\right)\cdot R^2}\\\\ &=V_3\cdot\frac{D}{D+D-D^2+1-D}=V_3\cdot\frac{D}{-D^2+D+1} \end{align*}$$
Which is the same thing Andy came up with. Shuffling things around, you get a quadratic and a solution:
$$\begin{align*} -V_O\cdot D^2 + (V_O-V_3)\cdot D + V_O&=0, ~~~~~\textrm{where }a=-V_O, b= V_O-V_3, c=V_O\\ \\ \therefore D &= \frac{-b\pm\sqrt{b^2-4 a c}}{2 a}\\\\ &=\frac{-\left(V_O-V_3\right)\pm\sqrt{\left(V_O-V_3\right)^2-4\cdot\left(-V_O\right)\cdot V_O}}{2\cdot\left(-V_O\right)} \\ \\ &=\frac{V_3-V_O\pm\sqrt{V_3^2-2\cdot V_O\cdot V_3+5\cdot V_O^2}}{-2\cdot V_O}\\ \\ &\approx -2.73996779, 0.364967794 \end{align*}$$
Clearly, the negative value isn't permitted here. So the answer is 36.5% (or rounding one digit further, 36%.)
Now you've got different ways of thinking about it. Pick one that sings in your mind better and go with it.