Question

Consider the three resistors R1=11 Ω, R2=29 Ω, and R3=77 Ω in the configuration shown in the figure. A potential difference ΔV=3.5 V is applied between A and B 

Part (a) Express the equivalent resistance Re of the combination of R2 and R3 in terms of R2 and R3.

Part (b) Express the total resistance R off the combination of all three resistors in terms of R1, R2 and R3

Part (c) Calculate the numerical value ofthe total resistance R in Ω.

Part (d) Express the current I on R1 in terms of the potential difference ΔV between A and B and the equivalent resistance R

Part (e) Calculate the numerical value of I in A. 

Part (f) Express the voltage across R2, ΔV2, in terms of ΔV, I, and RI

Part (g) Express the current I2 through R2 in terms ΔV2, and R2.

Part (h) Calculate the numerical value of I2 in A.

Part (i) Calculate the numerical value of I2 through R2

Answer 1

Please comment below if any part is not correct, since we need to express final answers in given terms, so there can be some mismatch in questions where expressions are asked. Also check that in Part (i) I have calculated value of I3, (as picture is cropped in that part, so let me know if you need anything else in that part).

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Part A.

R2 and R3 are in parallel, So their equivalent will be

1/Re = 1/R2 + 1/R3

Re = R2*R3/(R2 + R3)

Part B.

Now Re and R1 are in series, So total equivalent resistance will be

Req = R1 + Re

Req = R1 + (R2*R3)/(R2 + R3)

(Req = R, as per notations)

R = (R1*R2 + R1*R3 + R2*R3)/(R2 + R3)

Part C.

R = (11*29 + 11*77 + 29*77)/(29 + 77)

R = 32.07 Ohm

Part D.

Using Ohm's law:

V = ieq*R

ieq = V/R = 3.5/32.07 = 0.11 Amp

Now remember in resistors parallel combination voltage distribution in each part will be same and in series combination current distribution in each resistor will be same.

ieq = i1 = i23

i1 = V/R

Part E.

As per question notations, i1 = I

So,

I = V/R

I = 3.5/32.07

I = 0.11 Amp

Part F.

Voltage across R2 will be

V2 = V - V1

where

V1 = Voltage drop in R1 = I*R1

V2 = V - I*R1

Part G.

from Ohm's law

I2 = V2/R2

Part H

I2 = (3.5 - 0.11*11)/29

I2 = 0.079 Amp

Part I

I3 = I - I2

I3 = 0.11 - 0.079

I3 = 0.031 Amp

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