mple 18.3 Equivalent Resistance I Solve a problem involving both series and para
ID: 1547092 • Letter: M
Question
mple 18.3 Equivalent Resistance I Solve a problem involving both series and parallel 64 stors 6.0 n Problem Four resistors are connected as shown in Figure 8.0 4.0 18.11a (a) (a) Find the equivalent resistance between points a and c. (b) What is the current in each resistor if a 42 V battery is 8.0 n connected between a and c? Strategy Reduce the circuit steps, as shown in Figures 18.11 b and 18.11 c, using the sum rule for resistors in 2.0 12 n series and the reciprocal-sum rule for resistors in parallel. (b) W W Finding the currents is a matter of applying Ohm's law while working backwards through the diagrams. 14 n WAN (c) Figure 18.11 (Example 18.3) The four resistors shown in (a) can be reduced in steps to an equivalent 14 S resistor. Solution (a) Find the equivalent resistance of the circuit. The 8.0 2 and 4.0 2 resistors are in series, so use Reg R1 R2 8.0 2 4.0 2 12 2 the sum rule to find the equivalent resistance between a and b. The 6.0 and 3.0 resistors are in parallel so use 1 1 the reciprocal-sum rule to find the equivalent Req Ri R2 6.00 3.00 2.00 resistance between b and c (don't forget to invert!) Reg 2.0 2 In the new diagram, 18.11b, there are now two Req R1 R2 12 2 2.0 14 resistors in series. Combine them with the sum ruleExplanation / Answer
3 ohm and 6 ohm are in parallel.Resultant of these two is R =(3 x 6 ) / (3 + 6)
= 18 / 9 = 2 ohm
2 ohm and 4 ohm are in series.Resultant of these two is R ' = 2 ohm + 4 ohm = 6 ohm
R and R ' are in series.
Therefore equivalent resistance Req = R + R ' = 2 ohm + 6 ohm = 8 ohm
Emf E = 40 volt
Total current I = E / Req
= 40 / 8 = 5 A
I1:I2 = (1/3) : (1/6)
= (6/3):(6/6)
= 2 : 1
So, I1 = I ( 2/ (2+1) )
= 5 ( 2/ 3)
= 10/3 = 3.3333 A
I2 = I(1/(2+1))
= 5/ 3 = 1.666 A
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