Show that in a series combination of resistors the total resistance can never be

Question

Show that in a series combination of resistors the total resistance can never be less than that of the largest resistor whereas in a parallel combination the total resistance can never be more than that of the smallest. Referring to the circuit shown below, complete the associated table using Ohm's law and your knowledge of series and parallel circuits. Note that the numbers 1, 2, 3, and 4 in the table refer to the resistance R of, the current/through, and the voltage V across, the four resistors R_1, R_2, R_3, and R_4 respectively.

Explanation / Answer

two resistors are connected in parallel with battery (emf V) .

and suppose current through battery is I.

then I = V / Req

in parallel connection , PD across both resistance will be same and equal to battery.

and current will divide in resistors say i1 in R1 then i2 in R2.

and i1 + i2= I

V = i1R1 = i2R2

i1 = V/R1   and i2 = V/R2

i1 + i2 = I

V/R1 + V/R2 = V/Req

1/Req = 1/R1 + 1/R2

now Req will always be less R1 and R2.

for series.

current will be same through both resistor say i.

V = iReq

and PD across R1 will be iR1 and across R2 will be iR2.

and using KVL, V1 + V2 = V

iR1 + iR2 = iReq

Req = R1 + R2

so Req will always be greater than R1 and R2.

12.

R4 and R2+R3 are in parallel.

so PD across them will be same.

current in R2 + R3 is i2 then

i2 (R2 + R3) = 10

r3 = 18

i2 (18 + R2) =10

and i4R4 = 10

and PD across R2 = i2R2 = 4

so 18i2 + R2i2 = 10

18i2 + 4 = 10

i2 = 6/18 = 0.33 A

so i2 = i3 = 0.33 A

and i2 + i4 = I (curent through r1 and battery that is 0.50)

i2 + i4 = 0.5

i4 = 0.5 - 0.33 = 0.17 A

R2 = V2 / I2 = 4 / 0.33 = 12 ohm

R4 = V4 / I4 = 10 / 0.17 = 60 ohm

PD across R1 = 15 - 10 = 5 volt

R1 = V1/I1 = 5/0.5 = 10 ohm

R1 = 10 ohm, R2 = 12 ohm . R3 = 18 ohm , R4 = 60 ohm

I1 = 0.5 A . I2 = 0.33A, I3 = 0.33 A, I4 = 0.167 A

V1 = 5V , V2 = 4 v, V3 = 6V, V4 = 10 ohm

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