Voltage instead "regulates" how fast a motor can run: the maximum speed a motor can reach is the speed at which the motor generates a voltage (named "Counter-electromotive force") which is equal to the voltage it receives from battery (disregarding power losses and frictions for simplicity).
Voltage is the cause, current is the effect. So both sources provide a voltage on their terminals. In the voltage source the voltage on the output terminals has, or is intended to have, a constant value. Sometimes it's left to the internal resistance of the voltage source to determine what the final voltage will be, and sometimes the voltage at the terminals is measured and fed back to a ...
The reverse voltage is the voltage drop across the diode if the voltage at the cathode is more positive than the voltage at the anode (if you connect + to the cathode). This is usually much higher than the forward voltage. As with forward voltage, a current will flow if the connected voltage exceeds this value. This is called a "breakdown".
According to Ohm's law, resistance varies directly with voltage You should read this the other way. Voltage varies directly with current. "R" is the constant of proportionality telling how much it varies. If I add in a resistor to a circuit, the voltage decreases. If you have a resistor in a circuit, with a current flowing through it, there will be a voltage dropped across the resistor (as ...
If power is a constant, then, yes, current and voltage are inversely proportional since power is their product. Again, this has nothing to do with Ohm's Law. Ohm's law says that voltage and current are proportional because resistance is constant. This fact, however, has nothing to do with constant power.
A current source can certainly have a voltage across it. If the voltage across a current source is zero, then it is not delivering or absorbing any power. However, if the voltage across the source is not zero, then it is either sourcing or sinking power into the rest of the circuit.
Most, or maybe all, topologies could end up outside of common mode voltage ranges at some specific time. What is important is to understand under what conditions will you be outside of the common-mode voltage range when designing a circuit, and if so will the op-amp you choose still suffice for your application?
I realize connecting two different voltage sources in parallel is a contradiction (in an ideal circuit). But if I were to connect this in practice and measure the voltage across points A and B, what value of voltage would it show?
An intuitive way to look at is that all the voltage is dropped across two resistors, and since the resistors are the same, the voltage drop across each will be the same, each taking half.
Voltage has exactly the same problem: one terminal can only "have a voltage" when compared to another terminal. Voltage acts like distance: voltage and distance are double-ended measurements. Or in other words, one terminal in a circuit always has many different voltages at the same time, depending on where we place the other meter lead.
