IR drop across a resistance is voltage.
The letter I means current, and the letter R means resistance. Current times resistance, by Ohm's law is voltage.
IR Drop means voltage drop. As voltage drop across any resistance is product of current (I) passing through resistance and resistance value(R ) , it is often written as IR drop.
V=IR by ohms law. Voltage across the resistor is the product of current flowing and resistance of the conductor at constant temperature.
A: There are tables that qualify IR drops for wire lenght. All wire do offer resistance to current this current will cause directly a volatge drop according to the wire resistance so it can be measured to find the IR drop
Voltage drop across a circuit is IZ, where I is current and Z is impedance. In other words IZ = IR + jIX, where R is resistance and X is inductance
as the resistance/impedance of the the conducting wire is so small as compared to the load, there is a very low voltage drop across the conducting wires. keep the Ohm's law in mind i.e. V=IR <><><> Agree- but making the conducting wires LONG enough, and there will be enough resistance for a voltage drop.
In Ohm's Law, E stands for voltage, I stands for current (amps), and R stands for resistance. Ohm's Law states: E = IR or voltage equals current times resistance. This means that current flowing through a wire (that has resistance) produces a voltage drop in the wire. Since the voltage drop is the result of current flowing through a resistance, old-school engineers will sometimes refer to it as "IR drop". So, since E = IR, saying "IR" is the same as saying "voltage".
Resistance is decreases then the voltage drop across it is decreases because of ohms law (V=IR),due to this power can also be decreases because of P=VI. so finally we can say that resistance decreases ,power also decreases.
Yes
In a d.c. circuit, voltage drop is the product of resistance and current through that resistance.
Using ohm's law, V=IR then R=V/I =6/0.0015=4000 ohm = 4k ohm resistor.
Voltage drop is resultant of IR ie current and the line resistance, not dependent on impressed emf
You need to use the formula E = IR + Ir where: E is the e.m.f. of the power supply (the theoretical maximum voltage across the terminals when no current is flowing) I is current R is resistance of the circuit (load resistance) and r is the internal resistance of the power supply. Therefore, you can rearrange this formula to give r: E = IR + Ir (Subtract IR) E-IR=Ir (divide by I) (E-IR)/I=r or r=(E-IR)/I