Question

Calculate the resistance between points A and B (RAB) for the following resistor networks:

Answer 1

Given that,

According first figure (1),

$R_{1}=500\ \Omega$

$R_{2}=500\ \Omega$

$R_{3}=500\ \Omega$

$R_{4}=500\ \Omega$

All resistance are same.

So, 500 Ω and 500 Ω are connected in series.

(i). We need to calculate the equivalent resistance

Using formula of series

$R=R_{1}+R_{2}$

Put the value into the formula

$R=500+500$

$R=1000\ \Omega$

Now, 500 Ω and 500 Ω are connected in series.

We need to calculate the equivalent resistance

Using formula of series

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

Put the value into the formula

$R'=500+500$

$R'=1000\ \Omega$

Now, R and R' are connected in parallel

We need to calculate the resistance between points A and B

Using formula of parallel

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

$\dfrac{1}{R_{AB}}=\dfrac{1}{1000}+\dfrac{1}{1000}$

$\dfrac{1}{R_{AB}}=\dfrac{1}{500}$

$R_{AB}=500\ \Omega$

(ii). Given that,

According to figure (2),

$R_{1}=1\ k\Omega$

$R_{2}=1\ k\Omega$

$R_{3}=1\ k\Omega$

$R_{4}=1\ k\Omega$

All resistance are same.

Here, R₂ , R₃ and R₄ is connected in series

We need to calculate the equivalent resistance

Using series formula

$R=R_{2}+R_{3}+R_{4}$

Put the value into the formula

$R=1+1+1$

$R=3\ k\Omega$

We need to calculate the equivalent resistance between A and B

Using parallel formula

$\dfrac{1}{R_{AB}}=\dfrac{1}{R_{1}}+\dfrac{1}{R}$

Put the value into the formula

$\dfrac{1}{R_{AB}}=\dfrac{1}{1}+\dfrac{1}{3}$

$R_{AB}=\dfrac{3}{4}$

$R_{AB}=0.75\ k\Omega$

(iii) Given that,

According to figure (3),

$R_{1}=2\ k\Omega$

$R_{2}=100\ \Omega$

$R_{3}=5\ k\Omega$

$R_{4}=470\ \Omega$

Here, R₁ and R₃ are connected in parallel

We need to calculate the equivalent resistance

Using formula of parallel

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

$\dfrac{1}{R}=\dfrac{1}{2}+\dfrac{1}{5}$

$R=1.42\ k\Omega$

R₂ and R₄ are connected in parallel

We need to calculate the equivalent resistance

Using formula of parallel

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

$\dfrac{1}{R'}=\dfrac{1}{100}+\dfrac{1}{470}$

$R=82.45\ \Omega$

We need to calculate the equivalent resistance between A and B

Using formula of series

$R_{AB}=R+R'$

Put the value into the formula

$R_{AB}=1.42\times1000+82.45$

$R_{AB}=1502.45\ \Omega$

$R_{AB}=1.50\ k\Omega$

(iv). Given that,

According to figure (4),

$R_{1}=250\ \Omega$

$R_{2}=470\ \Omega$

$R_{3}=940\ \Omega$

The equivalent resistance between A and B is 940 Ω

Since, 250 Ω and 470 Ω a loop at node B.

So, We can ignore them.

(v). Given that,

According to figure (5),

$R_{1}=2.2\ k\Omega$

$R_{2}=2.2\ k\Omega$

$R_{3}=2.2\ k\Omega$

$R_{4}=2.2\ k\Omega$

Here, R₃ and R₄ are connected in series

We need to calculate the equivalent resistance

Using formula of series

$R=R_{3}+R_{4}$

Put the value into the formula

$R=2.2+2.2$

$R=4.4\ k\Omega$

Now,  R₁, R₂ and R are connected in parallel

We need to calculate the equivalent resistance between A and B

Using formula of parallel

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

Put the value into the formula

$\dfrac{1}{R'}=\dfrac{1}{2.2}+\dfrac{1}{2.2}+\dfrac{1}{4.4}$

$\dfrac{1}{R'}=\dfrac{22}{25}$

$R'=880\ \Omega$

(vi). Given that,

According to figure (4),

$R_{1}=220\ \Omega$

$R_{2}=330\ \Omega$

$R_{3}=470\ \Omega$

$R_{4}=100\ \Omega$

Here, 330 Ω and 470 Ω are connected in parallel

We need to calculate the equivalent resistance

Using formula of parallel

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

Put the value into the formula

$\dfrac{1}{R}=\dfrac{1}{330}+\dfrac{1}{470}$

$\dfrac{1}{R}=\dfrac{8}{1551}$

$R=193.87\ \Omega$

Now, 193.87 Ω and 220 are connected in series

We need to calculate the equivalent resistance

Using formula of series

$R'=R+R_{1}$

Put the value into the formula

$R'=193.87+220$

$R'=413.87\ \Omega$

Now, R₄ and R' are connected in parallel

We need to calculate the equivalent resistance between A and B

Using formula of parallel

$\dfrac{1}{R}=\dfrac{1}{R_{4}}+\dfrac{1}{R'}$

Put the value into the formula

$\dfrac{1}{R}=\dfrac{1}{100}+\dfrac{1}{413.87}$

$R_{AB}=80.53\ \Omega$

Hence, (i). The equivalent resistance between A and B is 500 Ω.

(ii). The equivalent resistance between A and B is 0.75 Ω.

(iii). The equivalent resistance between A and B is 1.50 kΩ.

(iv). The equivalent resistance between A and B is 940 Ω.

(v). The equivalent resistance between A and B is 880 Ω.

(iv). The equivalent resistance between A and B is 80.53 Ω.

Answer 2

Answer:

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