Where V=stands for voltage, I=stands for current and R=stands for resistance. voltage=current multiplied by resistance so, by transposition, resistance=voltage divided by current. or R=V/I, R=9/3 The resistance is 3 ohms.
as by ohm's law, v=ir
r = v/i
therefore the resistance is 9/3=3 ohms
Use Ohm's law. V = I * R where V is voltage in volts, I is current in amperes, and R is resistance in ohms.
I have no idea
You need to calculate the equivalent resistance. For instance, if the three resistors are connected in series, simply add all the resistance values up. Then, you calculate the current (in amperes) using Ohm's Law (V=IR); that is, you need to divide the voltage by the resistance.
No current flows through the battery. There is a current through the external circuit. I = E/R = 9/10 = 0.9 amperes.
The rating of 450 amp-hours in a battery bank means that the battery can supply 450 amperes for one hour, so to speak. In reality, these ratings are normalized to an eight hour rate, which means that this battery can supply about 56 amperes for eight hours. Actual run time at 450 amperes would be somewhat less than an hour.
Use Ohm's law. V = I * R where V is voltage in volts, I is current in amperes, and R is resistance in ohms.
Ohm's law: Voltage is resistance times current. 80 ohms time 0.5 amperes = 40 volts.
I have no idea
The apmeres depend on the resistance of the circuit. The battery will be a certain voltage, and dividing the voltage by the resistance gives you amperes. V = I*R
You need to calculate the equivalent resistance. For instance, if the three resistors are connected in series, simply add all the resistance values up. Then, you calculate the current (in amperes) using Ohm's Law (V=IR); that is, you need to divide the voltage by the resistance.
1.5 volts
A resistance of 3 ohms connected between the terminals of a 9-volt battery will result in a current of 3 Amperes. If the battery is one of those little ones with snaps on top, it may be able to produce 3 amperes of current for about 3 seconds before it rolls over and totally dies.
To find the net resistance of the circuit connected to the battery in the figure, you need to calculate the total resistance by adding up the individual resistances in the circuit.
some batterys have a line on them which shows how much they have left
2 amperes (current = voltage/resistance)
The current depends on the total effecvtive resistance of everything connectedacross the battery.If the bulb is the only component there, then the current is E/R = 9/12 = 0.75 amperes.
No current flows through the battery. There is a current through the external circuit. I = E/R = 9/10 = 0.9 amperes.