Question

consider the following network of six identical resistors as shown in figure.If the equivalent resistance of this circuit between A and B, B and C and A and C are RAB,RBC,RAC respc. then which of the following is true.

(a) Rab= Rbc > Rac
(B) Rac > Rbc > Rab
(C )Rbc > Rac > RAB
(d) Rab = Rbc = Rac​

Answer 1

Answer:

d) is correct answer..................

Answer 2

$R_{ab}>R_{bc}>R_{ac}$ is the correct answer.

Explanation:

According to given diagram,

Both R is connected in parallel between B and C

We need to calculate the resultant resistance R'

Using parallel formula

$\dfrac{1}{R'}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}$

Put the value into the formula

$\dfrac{1}{R'}=\dfrac{1}{R}+\dfrac{1}{R}$

$R'=\dfrac{R}{2}$

Here, three resistance are connected in parallel between A and C

We need to calculate the resultant resistance R''

Using parallel formula

$\dfrac{1}{R''}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$

Put the value into the formula

$\dfrac{1}{R''}=\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}$

$R''=\dfrac{R}{3}$

Now, Resistance R and R' are connected in series between A and C

We need to calculate the resultant resistance R'''

$R'''=R+R'$

Put the value intyo the formula

$R'''=R+\dfrac{R}{2}$

$R'''=\dfrac{3R}{2}$

Now, R''' and R'' are connected in parallel in between A and C

We need to calculate the resultant resistance between A and C

$\dfrac{1}{R_{ac}}=\dfrac{1}{R'''}+\dfrac{1}{R''}$

Put the value into the formula

$\dfrac{1}{R_{ac}}=\dfrac{2}{3R}+\dfrac{3}{R}$

$\dfrac{1}{R_{ac}}=\dfrac{11}{3R}$

$R_{ac}=\dfrac{3R}{11}$

Now, Resistance R'' and R' are connected in series between A and B

We need to calculate the resultant

$R''''=R''+R'$

Put the value into the formula

$R''''=\dfrac{R}{3}+\dfrac{R}{2}$    

$R''''=\dfrac{5R}{6}$

Now, R'''' and R are connected in parallel between A and B

We need to calculate the resultant resistance between A and B

$\dfrac{1}{R_{ab}}=\dfrac{1}{R''''}+\dfrac{1}{R}$

Put the value into the formula

$\dfrac{1}{R_{ab}}=\dfrac{6}{5R}+\dfrac{1}{R}$

$\dfrac{1}{R_{ab}}=\dfrac{11}{5R}$

$R_{ab}=\dfrac{5R}{11}$

Now, Resistance R'' and R are connected in series between B and C

We need to calculate the resultant

$R'''''=R''+R$

Put the value intyo the formula

$R'''''=\dfrac{R}{3}+R$

$R'''''=\dfrac{4R}{3}$

Now, R''''' and R' are connected in parallel between B and C

We need to calculate the resultant resistance between B and C

$\dfrac{1}{R_{bc}}=\dfrac{1}{R'''''}+\dfrac{1}{R'}$

Put the value into the formula

$\dfrac{1}{R_{bc}}=\dfrac{3}{4R}+\dfrac{2}{R}$

$\dfrac{1}{R_{bc}}=\dfrac{11}{4R}$

$R_{bc}=\dfrac{4R}{11}$

Hence, $R_{ab}>R_{bc}>R_{ac}$ is the correct answer.

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Topic : resultant resistance

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