5 ation could have been ratio would still be less than one and Requivalent WOuld

Question

5 ation could have been ratio would still be less than one and Requivalent WOuld have fors are placed in parallel, the equivalent resistance must be less than the smaller of the two ranged so that R, was outside the brackets and R1 was inside. In this been less than R2. In conclusion, if mportant consideration for this equation is when the two resistors in parallel differ very greatly in es. Analyzing the equation for this situation will show that the equivalent resistance will be ual to the value of the smaller of the two resistors. /Parallel Combinations own to the right is a circuit that ongous combination of resistors in parallel. Of interest in circuits like be the equivalent resistance of the t as well as the voltage across through each resistor. The first entify which resistors are initially which ones are in parallel and e the equivalent resistance of ations. Examination of the ow that resistors R7 and Rs are esistors initially in series with an istance of R1,8R7+ R8 R2 Rs R6 parallel combinations made of ith R3 as well as Rs in parallel Figure 2.5. Humongous Combo equivalent resistances will be: RL R5 R = R,+R, 2-3 and Rsj ssuch as Rs and Re can be confusing because they may not look to be "parallel." It may sliding the upper ends of those resistors toward each other along the connecting wire until they do appear parallel. You cannot slide a connection across a resistor but ng a wire.

Explanation / Answer

Given, all resistors are same, value = R = 1000 ohm

From the figure we can see

R2, R3 are in parallel, so Reff1= R/2 = 500 ohm

R5, R6 are in parallel, so Reff2 = R/2 = 500 ohm

R7 and R8 are in series, so Reff3 = 2R = 2000 ohm

Reff2 and Reff3 are in parallel, so Reff4 = (1/500 + 1/2000)^-1 = 400 ohm

Reff1, Reff4, R4 are in series, so Reff5 = 500 + 400 + 1000 = 1900 ohm

Reff5 and R1 are in parallel, Reff = (1/1900 + 1/1000)^-1 = 974.358 ohm

so, current leaving battery = i

from ohms law

V = iR

V = 12 V

so i = V/Reff = 0.012315 A

1. Voltage across R1 = 12 V

Current through R1 = 12/1000 = 0.012 A

2,3. Current through (R2, R3 comination) = 0.012315 - 0.012 = 0.000315 A

so current through R2,R3 = 1.579*10^-4 A ( current divides equally amongst equal resistors)

Voltage across R2, R3 = 1.579*10^-4*1000 = 0.1579 V

4. current through R4 = 0.00315 A

voltage across R4 = 0.00315*1000 = 3.15 V

5,6 Voltage acropss R5, R6 = 12 - 0.1579 - 3.15 = 8.6921

current through R5, R6 = 8.6921/1000 = 8.6921*10^-3 A

7, 8 voltage across R7,R8 = 8.6921/2 = 4.346 V

current through R7,R8 = 4.346/1000 = 4.346*10^-3 A

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