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A battery is rated to supply a certain number of volts. However, it actually supplies less, because they are "lost" as the current has to get out of the battery in the first place.(The battery has internal resistance)The amount of lost volts depends on the current being drawn:The less resistance a circuit has, the more current is drawn, because it's easier to flow.Example:If the circuit has little resistance, it draws a large current and the battery's internal resistance causes more lost volts.If the circuit has high resistance, it draws a small current and there are fewer lost volts.This is why when you short-circuit a battery (give it hardly any resistance to go through) it heats up and may explode. A large current is drawn and all the volts are used by the battery's internal resistance.
If two or more circuit elements are connected in series, the current must pass through each of them in turn. If two or more circuit elements are connected in parallel, that means there is a "fork in the road". In this case, part of the current will pass through one element, and part, through another one.
nothings because they is no wire connected to the circuit so no current can flow through
The idea is to use Ohm's Law (V=IR). In this case, the relevant voltage is the 10.2 V across the lamp.
I = E / R = 120 / 14 = 8.571 Amp. (rounded)
Your current will be 30/R Amps. Where R is the resistance in Ohms.
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.
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
V = IR Voltage = Current * Resistance so 9 = 0.25 * R Hence R =36 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.
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).
A shunt resistance is a low resistance connected parallel to the galvanometer so that a large portion of current passes through the low resistance and a small fraction of current passes through the galvanometer this saves the galvanometer from damage
V = I * R. 1.5 = 8*IThe current flow is 3/16 Amps.
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 !)
Because they have internal resistance. Current flow through this internal resistance produces heat, just like current flow through ordinary resistors does. The current can be from use of the battery or charging the battery (if it is rechargeable). Usually the internal resistance of a battery increases with age, meaning older batteries tend to run hotter than fresh ones.