0
It's the product of the resistance of that resistive load and the current passing through it.
Wiki User
∙ 10y ago
Add your answer:
Earn +20 pts
What is the formula for Ohms Law?
V = I * RThe equation you are looking for is R = V/I, although this is derived from the definition of the ohm, not Ohm's Law!Ohm's Law is merely a statement which, in effect, says that for a limited range of conductors, the ratio of voltage to current is constant for variations in voltage.If the ratio of voltage to current changes for variations in voltage, then Ohm's Law does not apply.However, the ratio of voltage to current will always tell you what the resistance of a load or device happens to be for that particular ratio.
What is the voltage of a battery if it is connected across an 8 ohm resistor and 0.75 amps of current flows through the resistor?
The battery has 6 volts across its terminals. The way to discover it is to apply Ohm's law. It (Ohm's law) comes in 3 "flavors" that look a bit different but all say exactly the same thing. Here they are: E = I x R [Voltage equals current times resistance.] I = E/R [Current equals voltage divided by resistance.] R = E/I [Resistance equals voltage divided by current.] In these equations, voltage is E, current is I and resistance is R. They are measured in units of volts, amperes (or amps) and ohms, respectively. In your problem, we have the resistance (R) and the current (I). We need to find the voltage (E), and the formula E = I x R is the logical choice to discover the voltage. As E = I x R here, E = 0.75 x 8 = 3/4 x 8 = 6 volts. Piece of cake.
If the resistance of an electric circuit is 12 ohms and the voltage in the circuit is 60 volts the current flowing through the circuit?
V = I times R where V = voltage, I = current and R = resistance. Further, I = V / R.As I = V / R, I = 60 /12 = 5 amps.V=IR , where V=60 volts R=12 ohms so I = V/R = 60/12 = 5 Amp.
What is RMS in electricity?
RMS stands for Root Mean Square. Power is calculated as V2/R where V is the voltage and R is the resistive component of a load, This is easy toi calculate for a DC voltage, but how to calculate it for a sinusoidal voltage? The answer is to take all the instantaneous voltages in the sine wave, square them, take the mean of the squares, then take the square root of the result. This is defined as the "heating effect voltage". For a sine wave, this is 0.707 of the peak voltage.
If the resistance in the circuit is increased what will happen to the current and voltage?
* resistance increases voltage. Adding more resistance to a circuit will alter the circuit pathway(s) and that change will force a change in voltage, current or both. Adding resistance will affect circuit voltage and current differently depending on whether that resistance is added in series or parallel. (In the question asked, it was not specified.) For a series circuit with one or more resistors, adding resistance in series will reduce total current and will reduce the voltage drop across each existing resistor. (Less current through a resistor means less voltage drop across it.) Total voltage in the circuit will remain the same. (The rule being that the total applied voltage is said to be dropped or felt across the circuit as a whole.) And the sum of the voltage drops in a series circuit is equal to the applied voltage, of course. If resistance is added in parallel to a circuit with one existing circuit resistor, total current in the circuit will increase, and the voltage across the added resistor will be the same as it for the one existing resistor and will be equal to the applied voltage. (The rule being that if only one resistor is in a circuit, hooking another resistor in parallel will have no effect on the voltage drop across or current flow through that single original resistor.) Hooking another resistor across one resistor in a series circuit that has two or more existing resistors will result in an increase in total current in the circuit, an increase in the voltage drop across the other resistors in the circuit, and a decrease in the voltage drop across the resistor across which the newly added resistor has been connected. The newly added resistor will, of course, have the same voltage drop as the resistor across which it is connected.
What is the voltage across the load R and the current through R when the switch is open in the following circuit?
There is insufficient information in the question to properly answer it. You did not provide details of the "following circuit". Please restate the question.
Why measure voltage with a meter and not just take it as 12V from the power pack?
The power pack has internal resistance of its own, equal to r. The voltage across the load resistance R is therefore: VR = 12(R / R + r) Thus the P.D. (VR) varies according to the load resistance R; it is impossible to ascribe an absolute value to it.
How galvanometer measure the exact voltage with series high resistor?
Assuming galvanometer has zero or negligible internal resistance. If u connect resistor R>>RL(Load resistance) and connect it parallel to RL, it will hardly cause any change in voltage across load resistance. Suppose small current Ig goes through galvanometer. Since galvanometer have zero internal resistance, Voltage across RL = Voltage across R = IgR
What size resistor needed to drop 3 volts dc to 2.5 volts dc?
You will need to take the resistance of the load into account if you are going to design a voltage divider. The resistance of the load can completely change the voltage ratio of a voltage divider if not factored into the calculation. you can measure or read R(load), then R(needed) = 0.8 R(load)
What is the law of the curve for power against voltage?
P = I x E (where E is voltage) so it's linear. ======================== The power dissipated by a component or circuit depends not only on the voltage across the load, but also on the characteristics of the load itself. If the resistance/impedance of the load is constant, then P = E2/R so it's proportional to the square of voltage.
Why is the formula squared in power dissipation of electrical resistance?
The fundamental equation for the power of any load is the product of the voltage across the load and the current through it: P = U I.Since voltage is the product of current and resistance (U = I R), we can substitute for voltage in the original equation:P = U I = (IR) I = I2R
Voltage across resistor is doubled the current is?
Ohm's law states that the voltage across a resistor is the product of the current times the Resistance or V=I x R (I times R). V is Voltage, R is Resistance, and I is Current or Amperage. So if the Voltage is doubled and Resistance stays the same, the Current will be doubled.
Why does the terminal voltage decreases as the load current increases?
A: That will happen anytime the voltage source is not able to provide the power needed for the load. If the load exceed the power available from the source the voltage will be reduced as IR drop from the source
How do you figure amperage from voltage 1500?
To answer this question the resistance of the load is needed. I = E/R.
Will voltage drop across a capacitor?
basically a capacitor will charge to the input DC level however it will mathematically never happen since capacitors charge at a certain rate the voltage drop across a capacitor will follow the R C time constant or 63% of the applied voltage for a unit time.AnswerIn the case of an a.c. supply, yes, there will be a voltage drop across a capacitor. In the case of an 'ideal' capacitor, this will be the product of the load current and the capacitive reactance of the capacitor.
Will high voltage give you high amps?
If the resistance of the circuit remains the same, yes. E = I x R Formulas for you to use are E/I = R, E being voltage and I being current or amps.AnswerNot necessarily. For example, electricity transmission is only possible because, for a given load, the higher the voltage, the lower the resulting load current. So the answer is that it depends on the load'!
How are the voltage drop across R and XL related in series RL circuit?
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
