V=IR
To calculate your voltage drop (V), you multiply your resistance (R, measured in Ohms) and current (I, measured in Amps [A]) by each other the. The number you are left with should be your voltage drop.
To measure the voltage drop in a circuit 1st we have to measure the each resistance, for which drop is occurring and than we have to measure current flowing through each resitance. After that we have to multiply current and resistance for each resistance which will ngive voltage drop in each resistance. Now to have total resistance we have to sum up all the voltage drops in whole circuit.
1 V = 1 J/C (1 joule per coulomb). That is, if there is a potential difference (or a "voltage") of one volt, you will require (or gain, depending on the sign) 1 joule of energy to move a charge of 1 coulomb across that potential difference.
for calculating the voltage drop in ckt first of all you have to calculate the total resistance of the ckt as well as total current also then applying ohm's law v=IR u can calculate the voltage drop across a ckt
While the answer is correct you might investigate thevenin theory.
1 kg m2 / (A s3)
Answer
A volt is defined as 'the potential difference between two points such that the energy used in conveying a charge of one coulomb from one point to the other is one joule'.
Since the voltage drop occurs across both conductors, we must take the resistance of both conductors in the cable into account. So, if the resistance of one conductor is R, then the total voltage drop along that cable will be:
Voltage drop = I x I x 2R
The output of the generator.
E=IxR
E=P/I
Yes it Does. V=I*R, I=V/R, R=V/I
it gives the reciprocal of resistance
A voltage of 13.8 kV is equal to 13,800 volts.
Electric power is not defined as current divided by voltage. Electric power (Watts) is equal to amps times voltage
If you know the voltage you can calculate the amps. . Ampere I = power P / voltage V .
Voltage is equal to amperage time resistance. V=IR Therefore, I'd say voltage times amperage is equal to amperage squared times resistance. VI=IIR Really there's no point in multiplying the two. However, if you were to divide voltage by amperage, you would have the resistance of the circuit. V/I=R
It depends on the type of three-phase system. If it's a three-wire system, then the phase voltage is numerically equal to the line voltage. If it's a four-wire system, then the phase voltage is numerically equal to the line voltage divided by 1.732 -in your example, this works out to be 5.77 V.
Power, in 'watts'.
The voltage is greater than the applied voltage, why?
If you are referring to the voltage after the rectifiers in a powersupply, it is due to the voltage drop across the rectifiers.
No. Voltage divided by resistance is equal to current.
The terminal voltage is equal to the supply voltage and there is zero current.
By Kirchhoff's Voltage Law, the sum of the voltage drops around the series circuit will equal the voltage applied to the circuit.
Zero. The sum of the voltage drops across all loads plus the voltage rises due to sources in a complete circuit must equal zero.
The source voltage.
Less than or equal to 1KV comes under Low voltage.
The batteries can be connected in parallel or in series. In parallel, good batteries of the same voltage will have a total voltage across them equal to the voltage across one of them. Those batteries in series will have a total voltage equal to the sum of the voltage of each of the batteries.
zero? the supply voltage? the supply voltage minus the individual coltage drops? the sum of the individual voltage drops? which one?
A voltage of 13.8 kV is equal to 13,800 volts.