This question can be answered using voltage dividers. Assume the power supply consists of a voltage source and a resistor. With no load, all of the voltage source's voltage is dissipated by the internal resistor of 15V. When there is a load, there are two resistors in series. To calculate the internal resistance:
1. I=V/R. You know the 600ohm resistor dissipated 13.7V. So that would mean a current of 13.7/600=22.8mA
2. If the 600ohm resistor dropped 13.7, kirchoff's voltage law would tell us the internal resistor dropped 15-13.7=1.3V.
3. R=V/I, Use the current to calculate the internal resistance. 1.3/22.8mA = 56.9ohms
CommentFurther to the above answer, a voltage-source's voltage is not 'dissipated by the internal resistance when on no load'. On no load, there is no current passing through the internal resistance, so no 'voltage dissipation' can takes plac -i.e. the non-load voltage is 15 V.
an ideal ammeter has zero or negligible resistance when this is connected in series no effective resistance would be added in the circuit so that the value of curret that we get is exactly of the circuit only. but when the ammeter is connected in parllel as it has zero resistance , the resistor to which it is connected in parllel gets shorted and due to his the effective resistance of the circuit is changed and so the effective current ... due to this the w=value measured by the ammeter would be different (incresed due to dec. in effective resistance)
The three electrical quantities are current voltage and resistance. Current is measured in amperes (A) and is the rate at which electricity flows through a conductor. Voltage is measured in volts (V) and is the electrical force pushing the current through the conductor. Resistance is measured in ohms () and is the opposition to the flow of current. Current - measured in amperes (A) Voltage - measured in volts (V) Resistance - measured in ohms ()
Resistance is measured in ohms.By Ohm's law, resistance is voltage divided by current, which is (joules per coulomb) divided by (coulombs per second), which is joules-seconds divided by coulombs squared. (It is easier to just say ohms.)
Short answer: yes. Most modern multimeters will not be damaged by external power when measuring resistance. But they will give erroneous readings. It is best to remove the power and disconnect the measured resistance from the larger circuit. A multimeter determines resistance by applying a small voltage, and measuring the resulting current. If the resistor has an external voltage source, then it will interfere with the measurement. Furthermore, if the resistance is connected to a larger circuit, then the resistance of this larger circuit will also be involved.
Ohm
Real-world batteries do not have zero internal resistance. When one connects a load (resistance) to a battery, current begins to flow and the open-circuit potential is divided between the battery's internal resistance and the resistance of the load. Thus, one will measure a lower voltage at the battery terminals when a load is connected, compared to no-load conditions.
what is the diference betwean calculated and maesured value
To a very small resistance so a mv can be measured as a function of amperes.
Resistance is measured in Ohms.Resistance is measured in Ohms.Resistance is measured in Ohms.Resistance is measured in Ohms.
an ideal ammeter has zero or negligible resistance when this is connected in series no effective resistance would be added in the circuit so that the value of curret that we get is exactly of the circuit only. but when the ammeter is connected in parllel as it has zero resistance , the resistor to which it is connected in parllel gets shorted and due to his the effective resistance of the circuit is changed and so the effective current ... due to this the w=value measured by the ammeter would be different (incresed due to dec. in effective resistance)
An ammeter does not have an 'output resistance'. It's important that its resistance is low so as not to add additional resistance into the circuit to which it is connected, otherwise the 'measured current' would be lower than the actual current.
No, a resistor isn't measured at all. A resistor has a quality called "resistance" - and that value is measured. Resistance is measured in Ohms.
-- Connect a source of known, small voltage across the ends of the unknown resistance. -- Measure the resulting current through the unknown resistance. -- Divide (small known voltage)/(measured current). The quotient is the formerly unknown resistance.
it is measured in Ohmmeter Ohms
Resistance is measured in ohms.
Electrical resistance is measured in Ohms.
if we know resistivity of copper i.e is very small (1.68×10−8)transposing the definition to make resistance the subject (Pouillet's law):R is the electrical resistance of a uniform specimen of the material (measured in ohms, Ω) is the length of the piece of material (measured in metres, m)A is the cross-sectional area of the specimen (measured in square metres, m²).