Call the total effective resistance 'R'. If the values of the individual parallel resistors are 'A', 'B', 'C', 'D' etc.,
then 1/R = (1/A) + (1/B) + (1/C) + (1/D) etc. Or, R = 1 divided by { (1/A) + (1/B) + (1/C) + (1/D) } The more resistors there are in parallel, the SMALLER the effective resistance becomes.
lower than the low resistance....
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.
For the individual resistor, the current is constant, regardless of any other resister that's attached to it in parallel. The current that results from all the resistors combined decreases as the resistance of one or more of the resistors increases.
5 A
-- The voltage doesn't change. -- If the second light bulb is identical to the first, then the total resistance drops by half. -- If they're not identical, then we have to know the details of both before we can calculate their combined effective resistance.
Greek Resistance happened in 1941.
the voltage across that resistor will increase if it is in series with the other resistors. the current through that resistor will increase if it is in parallel with the other resistors.
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.
For the individual resistor, the current is constant, regardless of any other resister that's attached to it in parallel. The current that results from all the resistors combined decreases as the resistance of one or more of the resistors increases.
That depends. For example, if the circuit is consisted of two resistors, 2 ohms each, the equivalent resistance (Req) of these two resistors in series is 4 ohms, and the Req of these two resistors in parallel is 1 ohm. If the same voltage is applied, say 4 V.power consumed in a resistance = V2/R.The parallel circuit: Power = 4 * 4 / 1 = 16 [W].The series circuit: Power = 4 * 4 / 4 = 4 [W].With everything else the same, a parallel circuit consumes more energy than a series circuit.Note that circuits of only simple resistors are discussed. You need to consider each circuit on its merit.================================AnswerIt depends. In both cases, the total energy expended will be the sum of the energies expended by each individual load.
5 A
Maybe blow the fuse or burn out the wiring. An ammeter has an extremely low resistance. connecting it across the resistance causes the resulting parallel resistance to be slightly lower than the resistance of the ammeter 1/Rt = 1/R + 1/R(ammeter)
The value and kind of resistors, connected in series or paralell, type of dc motor, what is the purpose, etc.
If it is connected in series with a circuit then it might raise the resistance too high and fail the system. Parallel connection is a circuit is probably the best bet you have.
-- The voltage doesn't change. -- If the second light bulb is identical to the first, then the total resistance drops by half. -- If they're not identical, then we have to know the details of both before we can calculate their combined effective resistance.
Greek Resistance happened in 1941.
The resistance will go up.
Midnight Resistance happened in 1990.