Using Ohms law ... E = I * R
Where
E is electromotive force in volts
R is resistance in Ohms
I is the current in amps (why "I", I don't know maybe some one else can edit this line)
E = 2 amps * 10 ohms = 20 volts
Your current will be 30/R Amps. Where R is the resistance in Ohms.
No current flows through the battery. There is a current through the external circuit. I = E/R = 9/10 = 0.9 amperes.
in voltmeter we have internal Resistance and connected in series , to current don't transfer in voltmeter , and we have internal resistance in ammeter and connected in parallel , to most current transfer through the ammeter.
If 3 identical 45-ohm resistors are connected in parallel, the net effective resistance of the bunch ...and the load seen by the battery ... is 15 ohms. The current supplied by the battery is60/15 = 4 Amperes.(This assumes that the battery is capable of supplying 4 amps at 60 volts, or 240 watts !)
V = I * R. 1.5 = 8*IThe current flow is 3/16 Amps.
Your current will be 30/R Amps. Where R is the resistance in Ohms.
The net resistance of the circuit connected to the battery in Figure 1 is the total resistance that the current encounters when flowing through the circuit. It is calculated by adding up the individual resistances of all the components in the circuit.
The formula you are looking for is R = E/I. Resistance is stated in ohms.
If a battery is "shorted", meaning that its terminals are connected together through a low resistance, high current flows in the connection and the battery becomes discharged very soon. It makes no difference whether any part of the battery is connected to ground.
The current flowing through the circuit can be calculated using Ohm's Law: I = V/R, where I is the current, V is the voltage (12 volts), and R is the resistance (25 ohms). Plugging in the values, the current would be 0.48 amperes.
it determines how well the current flows through the wires. ANSWER: When there is no outside power connected to it. But some power is necessary to read the resistance so the meter battery will supply the current necessary to measure the IR drop and translate that to resistance
No current flows through the battery. There is a current through the external circuit. I = E/R = 9/10 = 0.9 amperes.
A torch typically has its battery connected in series with the light bulb, creating a simple circuit. This means the electrical current flows from the battery through the bulb and back to the battery in a continuous loop, allowing the light to work.
in voltmeter we have internal Resistance and connected in series , to current don't transfer in voltmeter , and we have internal resistance in ammeter and connected in parallel , to most current transfer through the ammeter.
Ohm's Law: V = IR (voltage = current times resistance).Ohm's Law: V = IR (voltage = current times resistance).Ohm's Law: V = IR (voltage = current times resistance).Ohm's Law: V = IR (voltage = current times resistance).
V = IR Voltage = Current * Resistance so 9 = 0.25 * R Hence R =36 Ohms
The electric current in a circuit is caused by the flow of electrically charged particles, typically electrons, through the wires. When the battery is connected, it creates a potential difference (voltage) that pushes the electrons to move from the negative terminal to the positive terminal, creating a continuous flow of current in the circuit.