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My quick read of a few articles on Q, the reaction quotient, suggest that it's generally rather small. Thus $ln(Q)$ is negative, so $- T * ln(Q)$ is positive and the potential difference increases with temperature.

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To address the follow up question, which requires more characters than a comment allows: In a simple idealized view, the Fermi level is the top energy level in the solid occupied by electrons. In silicon with no doping it sits at mid-gap: the valance band is full, the conduction band empty. In a thought experiment, if you had two separate chunks of ...

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But why when I connect a volt-meter across the whole diode will I not see this built in potential? Because at equilibrium (V=0) the Fermi-level throughout the whole device is flat. A voltage will only exist when there is a gradient in Fermi-level over the component (i.e. a voltage drop). Edit Maybe a more visual explanation would help? Think of the ...

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Current is produced in a metal when the free electrons in the metals acquire a drift velocity due to an electric field. But when these free electrons travel through the metal, their path is hindered by other atoms and particles and their electomagnetic pull. More the resistance, higher is this hindrance and lesser is the drift velocity. Hence, the current ...

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The answer depends on the circumstances: how do you change the resistance? Both the drift velocity and the number of available charge carriers can be changed. In a basic Drude model for electrical transport both, $n$, the charge carrier density and $\tau$, the time between collisions determine the resistance: $$\mathbf{J} = \left( \frac{n q^2 \tau}{m} ... 1 Your understanding is correct. From V=IR if voltage stays the same while resistance is increased, the current should be decreased. But if you have heard of another equation I = \frac{Q}{t} If current(I) is increased and the charge Q is fixed (Charge is fixed if the power supply is from power cells like battery). Time will be decreased. Which means ... 0 Yes this is common practice for stepper motors only that the strategy is a little bit different. Instead of the voltage value one controls the time average of the voltage through a pulse-width modulation. This works because the coil works like an integrator:$$ i_L(t) = i_0+\frac1{L}\int_{0}^t v_L\left(\bar t\right) d\bar t $$v_L would be$$ v_L(t) = ...

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The work required to move a charge $q$ between two locations $a$ and $b$ with a voltage difference $V_{ab}$ is $$w=V_{ab}q.$$ Differentiating with respect to $q$, one obtains $$\frac{dw}{dq}=V_{ab},$$ which is how the book got the equation. Basically, the intuition is that voltage is "work per charge moved".

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Your reasoning is correct. If the solar cell is modelled as a voltage source $E$ in series with an internal resistance $r$ and the cell is connected to a load resistance $R_L$, the series current is given by Ohm's law: $$I = \frac{E}{r + R_L}$$ or $$E = (r + R_L)\cdot I$$ The output voltage $V$ of the solar cell is the voltage across the load ...

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