False. In a parallel circuit, each light bulb operates independently, so removing one bulb does not affect the brightness of the others. The remaining bulbs will continue to receive the same voltage as before, maintaining their brightness.
if we remove a resistor from the parallel connection the effective resistance value will be increased.
If you are referring to an electrical circuit, a series circuit is wired in such a way that if one object is removed from the circuit, the circuit is broken and everything within the circuit loses power. In a parallel circuit different components of the circuit can be removed without disabling power to the rest of the devices within the circuit.
In a series circuit, if one component is removed or defective, the circuit will be broken and no current will flow. In a parallel circuit, if one component is removed or defective, the current will simply bypass that component and continue to flow through the other branches.
No, the total resistance increases.
It is a series circuit. In a series circuit, the components are connected in a single path so if one lamp is removed, the circuit is broken and the other lamp will not receive electricity.
By connecting it to a circuit with the power still applied. Ohm meters work best when the component under test is removed from the circuit to eliminate parallel paths of current through other components.
Because circuit don't break unlike in series.That means rest of the component are getting the voltage across them.
To connect each object in a circuit to a cell separately, you would use a parallel circuit layout. This setup allows each object to have its own branch connected directly to the cell, ensuring that each object operates independently of the others in the circuit. This way, if one object fails or is removed, the others will continue to function.
wirewound you mean? seams pretty self explanitory, wires that are wound form a mag field.
Capacity to the equivalent of the battery is reduced - that is, you'll run out of charge quicker but the voltage delivered will still be the same.
It must (i) increase, or (ii) decrease, or (iii) stay the same. If (iii), there's no point in having it at all, so consider what happens to stage gain at the lowest frequencies, as the capacitor has less and less effect on the circuit.
assume the following configuration: battery connected to 2 parallel resistors with an ammeter in series with the battery... observe the current measurement ... remove one of the resistors .... observe the current again, it will have decreased: if the resistors were of equal value, the current will decrease to half of its original value when one of the resistors is removed. Mathematics: V=IR (V- voltage, I - current, R - resistance in a parallel circuit, R=(R1*R2)/(R1+R2) where R1 and R2 are the values of resistance of the resistors. Before removal- Ib=V*(R1+R2)/(R1*R2) After removal (assume R2 is removed)- Ia=V/R1 so Ia/Ib=(R1*R2)/(R1*(R1+R2)) or Ia=Ib*(R2/(R1+R2) if R1=R2 then Ia=Ib*R2/(2*R2) or Ia=Ib/2 so the current after is 1/2 of that before.