No, it is less. Use the formula:
1/R = 1/R1 + 1/R2 + 1/R3...
Where R is the total (equivalent) resistance for the parallel circuit,and R1, R2, etc. are the individual resistance.
No, it is less. Use the formula:
1/R = 1/R1 + 1/R2 + 1/R3...
Where R is the total (equivalent) resistance for the parallel circuit,and R1, R2, etc. are the individual resistance.
No, it is less. Use the formula:
1/R = 1/R1 + 1/R2 + 1/R3...
Where R is the total (equivalent) resistance for the parallel circuit,and R1, R2, etc. are the individual resistance.
No, it is less. Use the formula:
1/R = 1/R1 + 1/R2 + 1/R3...
Where R is the total (equivalent) resistance for the parallel circuit,and R1, R2, etc. are the individual resistance.
No, it is less. Use the formula:
1/R = 1/R1 + 1/R2 + 1/R3...
Where R is the total (equivalent) resistance for the parallel circuit,and R1, R2, etc. are the individual resistance.
3 ohms
if we remove a resistor from the parallel connection the effective resistance value will be increased.
in a parallel circuit resistance decreases increasing the current.
Placing another resistor in parallel to an existing resistor will lower the total resistance in the circuit. RParallel = 1 / Summationi=1toN (1 / Ri)
The total effecive resistance of several individual resistances in parallel is less than the smallest individual resistance, so in that sense I guess you'd have to say that the lowest resistance 'dominates' the character of the whole parallel circuit.
To find equivalent resistance when you have both parallel and series resistors, start simple and expand... Find the smallest part of the circuit, such as a pair of resistors in series or a pair of resistors in parallel, and compute the equivalent single resistor value. Repeat that process, effectively covering more and more of the circuit, until you arrive at a single resistance that is equivalent to the circuit. For resistors in series: RTOTAL = R1 + R2 For resistors in parallel: RTOTAL = R1R2/(R1+R2)
no
if not disconnected you will measure the resistance of the circuit in parallel with the resistor.
500 ohms.
if we remove a resistor from the parallel connection the effective resistance value will be increased.
in a parallel circuit resistance decreases increasing the current.
The total resistance of a set of resistors in parallel is found by adding up the reciprocals of the resistance values, and then taking the reciprocal of the total. By removing a resistor the total current will lower. If you short out the parallel circuit as suggested it will take out the fuse that should be protecting the circuit.AnswerShorting-out a resistor in a parallel circuit, will act to short out the entire circuit, therefore, significantly increasing, not lowering, the current! And, as the previous answer indicates, this short-circuit current will operate any protective devices, such as a fuse.In a parallel circuit current does not lower but it will be increase if shorting-out one resistor in the two resistor parallel circuit, the circuit will become very low resistive and the larger current will flow through the short path.
The current through each resistor is equal to the voltage across it divided by its resistance for series and parallel circuits.
Placing another resistor in parallel to an existing resistor will lower the total resistance in the circuit. RParallel = 1 / Summationi=1toN (1 / Ri)
It depends on what you want to accomplish. If you want to decrease the resistance in a circuit, you would place the box in parallel to some other resistor. If you want to increase the resistance in a circuit, you would place the box in series.
The total effecive resistance of several individual resistances in parallel is less than the smallest individual resistance, so in that sense I guess you'd have to say that the lowest resistance 'dominates' the character of the whole parallel circuit.
The one with the highest resistance (or impedance, if the voltage is not DC).
To find equivalent resistance when you have both parallel and series resistors, start simple and expand... Find the smallest part of the circuit, such as a pair of resistors in series or a pair of resistors in parallel, and compute the equivalent single resistor value. Repeat that process, effectively covering more and more of the circuit, until you arrive at a single resistance that is equivalent to the circuit. For resistors in series: RTOTAL = R1 + R2 For resistors in parallel: RTOTAL = R1R2/(R1+R2)