ࡱ> ^`]7 bjbjUU .:7|7| QlRRR844Gt"6667 /7 ;7GGGGGGG$JI jK'GiC7507C7C7'GW;66 GW;W;W;C7j.66GW;C7GW; W;GG6 }fuN R7GGG0GGJL8JLGW; Series and parallel resistors In this section we deal with the mathematics of more than one resistor in a series or parallel circuit. Two resistors in series (Figure 1) The current (I) flowing through R1 and R2 is the same and so the potential differences across them are V1 = IR1 and V2 = IR2 But using Kirchoff's second rule the total potential difference across them is V = V1 + V2 Therefore V = IR = IR1 + IR2 where R is the effective series resistance of the two resistors. So: Two resistors in parallel (Figure 2) The potential difference (V) across each of the two resistors is the same, and the current (I) flowing into junction A is equal to the sum of the currents in the two branches (Kirchoff's first rule) therefore: I = I1 + I2 But since V = I1R1 = I2R2 I = V/R = V/R1 + V/R2  where R is the effective resistance of the two resistors in parallel. Notice that two resistors in series always have a larger effective resistance than either of the two resistors on their own, while two in parallel always have a lower resistance. This means that connecting two or more resistors in parallel, such as the use of a mains adaptor, will increase the current drawn from a supply (Look at the section that deals with the bath with the two plug holes!).  For three resistors in series the combined resistance is:  and for three resistors in parallel it is:  Resistors in parallel an alterative formula The formula for two resistors in parallel may also be written as: N.B this version cannot be extended simply to cover the case of three resistors in parallel. The version for three resistors being: R = R1R2R3/[R1R2 + R1R3 + R2R3]  A further note on circuits The p.d. between the points A and B in the circuit in Figure 3 may be found by considering the ratio of the voltage drops in the resistors in each branch of the circuit. If the potential at C is zero then: Since the total resistance between D and C is 10W the potential difference across the 5  resistor = (5/10)x12 = 6V Therefore potential at A = 6V Potential drop through the 6 resistor = (6/9)x12 = 8V Therefore potential at B = 4V And so the potential difference between A and B = 6 - 4 = 2V Some other interesting resistor networks may be found in the following file:  HYPERLINK "Resistance_networks.doc" See: Advanced text/Electricity and magnetism/Current electricity/Resistance networks PAGE  PAGE 1 V1 V2 Figure 1 R1 R2 I Resistors in series: R = R1 + R2 Resistors in parallel: 1/R = 1/R1 + 1/R2 Example Problems 1. Calculate the resistance of the following combinations: (a) 100  and 50  in series (b) 100  and 50  in parallel (a) R = R1 + R2 = 100 + 50 = 150 W (b) 1/R = 1/R1 + 1/R2 = 1/100 + 1/50 = and so R = 33 W 2. Calculate the current flowing through the following when a p.d of 12V is applied across the ends: (a) 200  and 1000  in series (b) 200  and 1000  in parallel (a) resistance = 1200 W. Using I = V/R = 12/1200 = 0.01 A = 10 mA (b) resistance = 167 W. Using I = V/R = 12/167 = 0.072 A = 72 mA 3. You are given one 100  resistor and two 50  resistors. How would you connect any combination of them to give a combined resistance of: (a) 200  (b) 125  (a) 100 W in series with both the 50 W . (b) the two 50 W in parallel and this in series with the 100 W R = R1 + R2 + R3 R = R1R2/[R1 + R2] 1/R = 1/R1 + 1/R2 + 1/R3 Example problems 1. Calculate the combined resistance of a 50W and a 100W resistor connected first in (a) series and then (b) in parallel (a) Series resistance = 50 + 100 = 150 W (b) Parallel resistance = 50x100/[50+100] = 5000/150 = 33.3 W 2. Calculate the combined resistance of two 50W resistors connected in parallel, the combination being joined in series to a further 50W resistor. 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