You describe a series circuit. This answer is for a series circuit. The total impedance is 16 Ohms, 12 + 4. The current by Ohm's Law (Amps = Volts / Ohms) is 40 / 16, or 2.5 Amps. Kirchoff's Current Law says that the current at every point in a series circuit is the same, so the current flowing into the 1st device, the 12 Ohm resistor, is 2.5 Amps.
Ohm's law is V = I * R.
The current I = V / R = 40 volts divided by 4 ohms = 10 amps.
in series: 2.5A.
I = E/R (Ohm's Law)
I = 40v / (12+4 = 16)
I = 40/16
I = 2.5.
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An ammeter has to measure to current flowing through the circuit. Resistance offers an obstruction to the current flow. So, if the resistance of an ammeter is large , the current measured by the ammeter will be quite less as compared to the actual amount of current flowing through the circuit which is undesirable. If ammeter has zero resistance , then it will give the exact value of current. But this is not practically possible because every material has some value of internal resistance which we can't control. For this reason , ammeter must have small resistance
There is a simple equation relating voltage (properly potential difference), current and resistance: V=IR Where V=potential difference, I=current and R=resistance So to answer: I=60/12 I=5
If the two 5 ohm resistors were in series, then the current would be 1.2 amperes. If they were in parallel, then the current would be 4.8 amperes. Ohm's Law: Current = Voltage divided by Resistance RSeries = Summation1toN RN RPARALLEL = 1 / Summation1toN (1 / RN)
no.because current always try to flow trough low resistance path.as short circuit has low resistance current pass trough short circuit
The current depends on the total effecvtive resistance of everything connectedacross the battery.If the resistor is the only component there, then the current is E/R = 12/3 = 4 amperes.
lowers the amt of electrical current flowing through it.
Depends - in the real world as a resistor gets hotter (current flowing through it) its resistance increases.
With the same voltage and resistance the current will be the same value.
Because there is many path for flowing current through circuit.
..using the formula Voltage(V)=Current(I) * Resistance(R) .. we can get the result ...current will be 5 Ampere
An ammeter has to measure to current flowing through the circuit. Resistance offers an obstruction to the current flow. So, if the resistance of an ammeter is large , the current measured by the ammeter will be quite less as compared to the actual amount of current flowing through the circuit which is undesirable. If ammeter has zero resistance , then it will give the exact value of current. But this is not practically possible because every material has some value of internal resistance which we can't control. For this reason , ammeter must have small resistance
Volts = Current x Resistance. The voltage is where the potential resides for the amount of current flowing through a resistance. Think about the voltage as a potential source of electrons that then flow through a circuit depending on the Load, or resistance in this example.
It isn't. If you're using superposition, you open circuit current sources and short voltage sources; this is because the current source declares the current that will be flowing through that branch. Both current and voltage sources have a finite internal resistance.
Resistance
Current flowing through a device depends on resistance offered by that device.
A Galvanometer can be used to detect the presence of current in a circuit. An ammeter can be used to know the magnitude of the current flowing through the circuit.
I=V/R Which means: amps(current) = voltage divided by resistance. 20= V/20