According to ohm's law V = IR.
So here
I=V/R
=12/4
I=3A
To determine which resistor had the smallest current through it when the voltage (v) was 12V, we can apply Ohm's Law (I = V/R). The resistor with the highest resistance (R) will have the smallest current (I) for a given voltage. Therefore, if you know the values of the resistors in the circuit, the one with the highest resistance will have the smallest current flowing through it when 12V is applied.
V=i*r=2*6=12v
Voltage can be calculated using Ohm's Law:Voltage = Current (A) x Resistance (Ω)Voltage = 4A x 3Ω = 12 VoltsTherefore, the battery is a 12 Volts.The power dissipated is Voltage x CurrentPower = 4A x 12V = 48 Watts
That would also depend on battery chemistry.
011.1
To determine which resistor had the smallest current through it when the voltage (v) was 12V, we can apply Ohm's Law (I = V/R). The resistor with the highest resistance (R) will have the smallest current (I) for a given voltage. Therefore, if you know the values of the resistors in the circuit, the one with the highest resistance will have the smallest current flowing through it when 12V is applied.
V=i*r=2*6=12v
The resistance can be calculated using Ohm's Law, which states that resistance (R) equals voltage (V) divided by current (I). In this case, R = 12V / 5A = 2.4 ohms.
Resistance = voltage / current = 12V / 2.5mA = 12V / (2.5 x 10-3 A) = 4.8 x 103 ohm
According to ohms law (R=V/I) if voltage increases the resistance also increases .For example: If voltage (V) becomes 2 times the resistance (R) also increases becomes 2 times keeping the current (I) same
Wattage or power rating of a product can be calculated by multiplying voltage rating and current rating. (Power = Voltage x Current). e.g. if device is working at 12V and 2A is the current rating. It is 24Watt. Since Voltage = Current x Resistance , for a resistive load power can also be calculated by Power = Current x Current x Resistance = I^2 x R = I square R
Using Ohm's Law (V = I * R), where V is voltage, I is current, and R is resistance, we can rearrange the formula to solve for resistance: R = V / I. In this case, R = 12V / 0.5A = 24 ohms. So the resistance of the electromagnet is 24 ohms.
The current produced by a 1.5V flashlight cell would be lower than the current produced by a 12V car battery. This is because current is directly proportional to voltage in a circuit with constant resistance, following Ohm's Law (V = IR). Therefore, the higher voltage of the 12V car battery would result in a higher current compared to the 1.5V flashlight cell.
The power dissipated by a circuit with a voltage of 12V and a current of 3A is 36W. Watts is Volts times Amps.
The output voltage of 12V 5W solar panel is 12V. The current is 0,41667A. These values are normaly calculated with 1000W/m2 of solar radiation.
Yes, you can use a 12V 5A output for a device that requires 12V 3.0A. The device will only draw the current it needs, so having a higher current rating in the power supply is fine. Just make sure the voltage matches and the polarity is correct.
Its normal. There's two components to electrical power: voltage and current. A voltage is merely a "potential" to do electrical work. The amount of resistance in a circuit with respect to the amount of potential is what determines how much current flows. Its simple ohms law: Voltage = Current * Resistance, Power = Voltage * Current. Pulling the fuses out of the fuse box leaves an open circuit, which is an infinite amount of resistance. So by pulling your fuses, you prevent current from flowing, but you haven't taken away your battery's potential to do work (its voltage.) The voltage is always there as long as the battery is charged, but no current will flow or power be realized unless you reduce that resistance to something less than infinite by connecting a circuit to it (or in this case, completing the circuit again by replacing the fuses.)