Voltage drop due to the resistance present in the series circuit causes voltage split over a series circuit.
voltage
Use a multimeter and test if there is a voltage drop or current over the wire circuit.
current
voltage over resistance/impredance. V / I * R V= Voltage I= Current R= Resistance or Impedance Cover what you want to know and the way the other two are laid out gives you your equation.
No, not directly. The supply voltage has to rise or the resistance has to fall to get over-current. If there was a secondary control voltage that was part of a voltage control circuit for a higher voltage, it is conceivable that a voltage drop in control circuit could cause an over-voltage in the supply. Motors are constant power devices, so this could be true for a motor. If you have a 1hp motor (loaded at 1hp), it will want to draw 1hp of power no matter the supply voltage. If the voltage dips, the motor will require more current to keep it spinning at it's normal speed.
voltage
Most likely a short circuit will cause no voltage. Due to the high current on a short circuit fault the over current protection of the circuit will trip. This will cut the voltage supply off completely.
The effective voltage of an electrical circuit is the measure of the average voltage over a complete cycle of alternating current. It is also known as the root mean square (RMS) voltage.
mcb is master circuit breaker. it is used in the circuit for the protection of equipments from over voltage.
Use a multimeter and test if there is a voltage drop or current over the wire circuit.
The voltage and current graph in a circuit shows the relationship between voltage (electrical potential) and current (flow of electricity) over time. It helps to understand how these two factors interact and affect each other in the circuit.
Yes. Most microprocessor based relays that have both voltage and current inputs can provide overcurrent and over voltage protection simultaneously. Short circuit current is the same as overcurrent.
current
voltage over resistance/impredance. V / I * R V= Voltage I= Current R= Resistance or Impedance Cover what you want to know and the way the other two are laid out gives you your equation.
What happens to the current in a circuit as a capacitor charges depends on the circuit. As a capacitor charges, the voltage drop across it increases. In a typical circuit with a constant voltage source and a resistor charging the capacitor, then the current in the circuit will decrease logarithmically over time as the capacitor charges, with the end result that the current is zero, and the voltage across the capacitor is the same as the voltage source.
No, not directly. The supply voltage has to rise or the resistance has to fall to get over-current. If there was a secondary control voltage that was part of a voltage control circuit for a higher voltage, it is conceivable that a voltage drop in control circuit could cause an over-voltage in the supply. Motors are constant power devices, so this could be true for a motor. If you have a 1hp motor (loaded at 1hp), it will want to draw 1hp of power no matter the supply voltage. If the voltage dips, the motor will require more current to keep it spinning at it's normal speed.
That depends. For example, if the circuit is consisted of two resistors, 2 ohms each, the equivalent resistance (Req) of these two resistors in series is 4 ohms, and the Req of these two resistors in parallel is 1 ohm. If the same voltage is applied, say 4 V.power consumed in a resistance = V2/R.The parallel circuit: Power = 4 * 4 / 1 = 16 [W].The series circuit: Power = 4 * 4 / 4 = 4 [W].With everything else the same, a parallel circuit consumes more energy than a series circuit.Note that circuits of only simple resistors are discussed. You need to consider each circuit on its merit.================================AnswerIt depends. In both cases, the total energy expended will be the sum of the energies expended by each individual load.