In this example we will use the concept of equivalent resistance to determine th

Question

In this example we will use the concept of equivalent resistance to determine the current in a circuit. Three identical resistors with resistances of 6.0 are connected as shown in (Figure 1) to a battery with an emf of 18.0 V and zero internal resistance. (a) Find the equivalent resistance of the resistor network. (b) Find the current in each resistor.

A)What is the current through R1 in the network in this example if the resistors aren’t equal, but instead R1=1.0, R2=5.0, and R3=3.0?

B)What is the current through R2 in the network in this example if the resistors aren’t equal, but instead R1=1.0, R2=5.0, and R3=3.0?

C)What is the current through R3 in the network in this example if the resistors aren’t equal, but instead R1=1.0, R2=5.0, and R3=3.0?

Explanation / Answer

a. Resistances in series are added algebraically
Resistances in parallel are added in a special way
Reff = (1/R1 + 1/R2)^-1

now in the given circuit
R3 and R2 are in parallel, both of them are in series with R1
so, Reff = R1 + (1/R2 + 1/R3)^-1 = 6 + (1/6 + 1/6)^-1 = 6 + 3 = 9 ohm
b. Current leaving the battery = i
from ohm's law
Reff*i = V
i = 18/9 = 2 A

current through R1 = 2 A

Now, Voltage drop across R1 = iR1 = 2*6 = 12 V
So voltage drop across R2 = V - 12 = 6 V
current through R2 = i2
V - 12 = i2*R2 = 6
i2 = 6/6 = 1A
current through R2 = 1 A

current through R3 = i - i2 = 1 A

A. if resistances arent equal
reff = R1 + (1/R2 + !/R3)^-1 = 1 + (1/5 + 1/3)^-1 = 2.875 ohm
current leaving battery = i = V/Reff = 6.26 A
current through R1 , i = 6.26 A
B. let current through R2 = I
then (i - I)R3 = I*R2
(i - I)3 = I*5
8I = 3i
I = 2.347 A

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