10kiloohm resistance with 10% tolerance and 20 kiloohm with 20% tolerance connected in series in first case and in parallel in second case. how to calculate total tolerance of the electrical circuit in each case
2006-10-17 05:20:25 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Engineering
In series, you sum all the impedances and determine the weighed average of the impedance tolerances. For your example:
R = R1+R2
R = 10 kΩ+20 kΩ = 30 kΩ
T = (T1*R1+T2*R2)/(R1+R2)
T = (10%*10 kΩ+20%*20 kΩ)/(10 kΩ+20 kΩ)
T = (5 kΩ)/(30 kΩ)
T = 16.7% tolerance
In parallel, you sum all the admittances and determine the weighed average of the admittance tolerances. For your example:
R = R1║R2 = (R1*R2)/(R1+R2)
R = (10 kΩ*20 kΩ)/(10 kΩ+20 kΩ) = 6.67 kΩ
T = (T1/R1+T2/R2)*(R1║R2)
T = (T1/R1+T2/R2)*(R1*R2)/(R1+R2)
T = (10%/10 kΩ+20%/20 kΩ)*(10 kΩ*20 kΩ)/(10 kΩ+20 kΩ)
T = (1%+1%)*(200)/(30)
T = 13.3% tolerance
2006-10-17 07:00:08 · answer #1 · answered by Deep Thought 5 · 2⤊ 0⤋
Tolerance Calculation Formula
2016-12-17 15:43:48 · answer #2 · answered by eichelberger 3 · 0⤊ 0⤋
the priority could nicely be broken down into 3 resistances all in sequence. The 0.2 ohm battery A the 0.a million ohm battery B and the unknown ohm resistance of the gentle bulb. understanding that the full contemporary I = 0.5A and the full voltage would be 3.0V because the two a million.5 V batteries are in sequence the full power of the circuit would be P = V*I P = 3.0 * 0.5 = a million.5W because P = R*I^2 the full resistance of the circuit will equivalent the full power divided via the sq. of the full contemporary Rt = Pt / It^2 for this reason R = a million.5 / (0.5^2) = 6 ohms understanding that Rt = R1 +R2 + R3 .... etc for a chain RL = 6 - 0.2 - 0.a million = 5.7 ohms and the ability dissipated via the batteries would be equivalent to (RA + RB) / Rt power disipated via batteries = (0.2 + 0.a million) / 6 = a million/20 to Recap finished Resistance = 6 ohms Resistance of Ligh bulb = 5.7 ohms finished power of circuit = a million.5 W Fraction of power dissipated via betteries = a million/20
2016-12-13 09:55:26 · answer #3 · answered by ? 4 · 0⤊ 0⤋
Hand select the resisters so that one is high the other is low and puts u in the 1% range.
2006-10-17 08:40:56 · answer #4 · answered by JOHNNIE B 7 · 0⤊ 1⤋
It depends
2016-08-23 08:57:10 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Really not sure about this one
2016-08-08 17:22:44 · answer #6 · answered by ? 3 · 0⤊ 0⤋