This looks like V=IR to me
Transposing to R = V/I and assuming that's a 12 ohm bulb
Let x be the unknown part of the circuit's resistance
12 + x = 8 / 0.62, so
12 + x = 12.9, therefore
x = 0.9 ohms
.x / 4 = 0.225 ohms
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
Current source means current generator for a circuit. An ideal current source gives all current to the circuit, but practically a current source does n't give all current to the circuit, instead, a source resistor is connected in parallel to the current source to indicate the current drop.
The total resistance of a circuit is the sum of the supply's internal resistance and its load resistance, because they are in series with each other. This is true regardless of the magnitude of, or the variation in, the current.
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.
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
Internal resistance. The ideal current source has no internal resistance in parallel with it (if it was set to supply no current it would act as an open circuit), and all the current it supplied would have to flow through its load (even if the load was an open circuit, in which case the voltage across the current source would be infinite). A real current source has the practical limitation that it must have an internal resistance in parallel with it, therefor some of the current it supplied is bypassed through that internal resistance and never reaches the load (if the load was an open circuit, then all the current supplied is bypassed and the resulting voltage drop across the internal resistance limits the voltage across the current source).
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.
internal resistance is always infinite in ideal current source .the internal resistance is in shunt with current source
Current source means current generator for a circuit. An ideal current source gives all current to the circuit, but practically a current source does n't give all current to the circuit, instead, a source resistor is connected in parallel to the current source to indicate the current drop.
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.
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.
All cells have internal resistance. When connected to a load, the resulting load current results in an internal voltage drop across the internal resistance. This voltage drop acts in the opposite sense to the cell's e.m.f., thus causing its terminal voltage to fall below that of the e.m.f. The greater the load current, the greater the difference between the terminal voltage and the e.m.f.
Doesn't work like that. Current drain is dependent on the (internal resistance of the battery and the) resistance/power requirements of what's connected to the battery. If shorted out, the current - unless the battery is fused or otherwise protected - can go into tens of amps.
That depends on the resistance connected. Use Ohm's Law: V=IR. Solving for current: I = V/R. If nothing is connected, there will be no current (infinite resistance).
The total resistance of a circuit is the sum of the supply's internal resistance and its load resistance, because they are in series with each other. This is true regardless of the magnitude of, or the variation in, the current.