as the given cells have the same current flowing in through them (current flowing through the cells connected in series is equal to the current flowing when connected in parallel ) equate the formula's of cells connected in series and cells connected in parallel.thus by equating we get the value of the internal resistor as 2 ohms.
Nothing "creates" energy. The chemical energy in gasoline can be converted to electrical energy by burning the gasoline in an internal combustion engine that's connected to a generator.
internal environment
Intrusive igneous rocks can have chemically and mineralogically identical counterparts in their extrusive igneous equivalent. The only difference between the two rocks would be their method of formation and texture. Example: granite and rhyolite from the same source of magma.
The major internal cause of slope failure is gravity.
Internal temperature is the temperature at or near the center of an object. The internal temperature is very important in cooking things like roasts, chickens, turkeys, etc. There are situations where outside temperature may be quite warm, but internal temperature may be quite cool. Things are to be cooked until the internal temperature is hot, often for some period of time.
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.
Internal resistance is approximately equal to 94.667
The flowing of electricity (amperage) is governed by the internal resistance of the connected device.
The plug itself will have a infinity value of resistance. If a load is connected to the plug, the internal resistance of the device will be present at the blades.
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=22.8mA2. 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 = 56.9ohmsCommentFurther 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.
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.
It depends on the application. Voltmeters have a high internal resistance, while ammeters have a low internal resistance.
That will depend on the sum of the load resistance and the internal resistance of the battery (this is true for all power sources, not just 6 volt batteries). Small compact batteries tend to have higher internal resistance and therefore are more limited in the current they can deliver to a given load than larger batteries.
The behaviour you are describing is, in fact, due to the internal resistance of the voltage source.When a voltage source, such as a battery or generator, is not connected to a load, its potential difference is simply the electromotive force (or 'no-load voltage') of that source. When a load is connected, a load current flows not only through the load itself, but also through the internal resistance of the source. This causes an internal voltage drop across the internal resistance, which acts in the opposite sense (i.e. in accordance with Kirchhoff's Voltage Law), or direction, to the electromotive force, thus reducing the voltage available at the terminals. The greater the load (i.e. the lower the load resistance), the greater the resulting load current, and the greater the internal voltage drop -and the lower the terminal voltage.
No, it is desirable for a battery to have a low internal resistance.
The internal resistance of a device is the resistance in ohms of that device. It is the resistance electrons need to overcome before electricity is said to flow.
To solve any D.C. circuit by using Thevenin Theorem,First of all load resistance RL is disconnected from the circuit and open circuit voltage across the circuit is calculated (known as Thevenin equivalent voltage)Secondly, the battery is removed by leaving behind its internal resistance. Now we calculate equivqlent resistance of the circuit ( called Thevenin equivalent resistance).Now we connect Thevenin Voltage in series with Equivalent resistance of the circuit and now connect load resistance across this circuit to calculate current flowing through the load resistance.Whereas in the case of using Norton theorem, we again remove the load resistance if any, and then short circuit these open terminals and calculate short circuit current Isc.Second step is same as in Thevenin theorem i.e. remove all sources of emf by replacing their internal resistances and calculate equivqalent resistance of the circuit.Lastly, join short circuit current source in parallel with equivalent resistance of the circuit. Now, we can calculate votage across the resistance which was connected in parallel with Isc.So, by knowing the open circuit voltage, we can calculate current flowing the resistance and on the other hand , by knowing the short curcuit current , we can calculate voltage across the resistance.