Question

Find the total resistance between A and B of the surface
R1=R2=10
R3=R4=20
R5=R6=5
R7=R8=40​

Answer 1

Find attachment carefully for the entire circuit and it's modification .

We need to find eq. resistance b/w A and B .

R₁ , R₂ are connected in parallel ,

$:\to \sf \dfrac{1}{R_{12}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

$:\to \sf \dfrac{1}{R_{12}}=\dfrac{1}{10}+\dfrac{1}{10}$

$:\to \sf \dfrac{1}{R_{12}}=\dfrac{2}{10}$

$:\to \bf R_{12}=5\ \Omega$

R₃ , R₄ are connected in parallel ,

$:\to \sf \dfrac{1}{R_{34}}=\dfrac{1}{R_3}+\dfrac{1}{R_4}$

$:\to \sf \dfrac{1}{R_{34}}=\dfrac{1}{20}+\dfrac{1}{20}$

$:\to \sf \dfrac{1}{R_{34}}=\dfrac{2}{20}$

$:\to \bf R_{34}=10\ \Omega$

R₅ , R₆ are connected in parallel ,

$:\to \sf \dfrac{1}{R_{56}}=\dfrac{1}{R_5}+\dfrac{1}{R_6}$

$:\to \sf \dfrac{1}{R_{56}}=\dfrac{1}{5}+\dfrac{1}{5}$

$:\to \bf R_{56}=\dfrac{5}{2}\ \Omega$

R₇ , R₈ are connected in parallel ,

$:\to \sf \dfrac{1}{R_{78}}=\dfrac{1}{R_7}+\dfrac{1}{R_8}$

$:\to \sf \dfrac{1}{R_{78}}=\dfrac{1}{40}+\dfrac{1}{40}$

$:\to \bf R_{78}=20\ \Omega$

Now R₁₂ , R₃₄ are connected in series ,

$:\to \sf R_{1234}=R_{12}+R_{34}$

$:\to \sf R_{1234}=5+10$

$:\to \bf R_{1234}=15\ \Omega$

R₅₆ , R₇₈ are connected in series ,

$:\to \sf R_{5678}=R_{56}+R_{78}$

$:\to \sf R_{5678}=\dfrac{5}{2}+20$

$:\to \bf R_{5678}=\dfrac{45}{2}\ \Omega$

Now R₁₂₃₄ , R₅₆₇₈ are connected in parallel ,

$:\to \sf \dfrac{1}{R_{12345678}}=\dfrac{1}{R_{1234}}+\dfrac{1}{R_{5678}}$

$:\to \sf \dfrac{1}{R_{eq}}=\dfrac{1}{15}+\dfrac{1}{\frac{45}{2}}$

$:\to \sf \dfrac{1}{R_{eq}}=\dfrac{1}{15}+\dfrac{2}{45}$

$:\to \sf \dfrac{1}{R_{eq}}=\dfrac{3+2}{45}$

$:\to \sf \dfrac{1}{R_{eq}}=\dfrac{5}{45}$

$:\to \sf \dfrac{1}{R_{eq}}=\dfrac{1}{9}$

$:\leadsto \textsf{\textbf{R _{\text{eq}} =9\ \boldsymbol \Omega }}\ \; \pink{\bigstar}$

Total resistance b/w A and B is 9 Ω .

Answer 2

Answer:

Find attachment carefully for the entire circuit and it's modification .

We need to find eq. resistance b/w A and B .

R₁ , R₂ are connected in parallel ,

:\to \sf \dfrac{1}{R_{12}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}:→

R

12

1

=

R

1

1

+

R

2

1

:\to \sf \dfrac{1}{R_{12}}=\dfrac{1}{10}+\dfrac{1}{10}:→

R

12

1

=

10

1

+

10

1

:\to \sf \dfrac{1}{R_{12}}=\dfrac{2}{10}:→

R

12

1

=

10

2

:\to \bf R_{12}=5\ \Omega:→R

12

=5 Ω

R₃ , R₄ are connected in parallel ,

:\to \sf \dfrac{1}{R_{34}}=\dfrac{1}{R_3}+\dfrac{1}{R_4}:→

R

34

1

=

R

3

1

+

R

4

1

:\to \sf \dfrac{1}{R_{34}}=\dfrac{1}{20}+\dfrac{1}{20}:→

R

34

1

=

20

1

+

20

1

:\to \sf \dfrac{1}{R_{34}}=\dfrac{2}{20}:→

R

34

1

=

20

2

:\to \bf R_{34}=10\ \Omega:→R

34

=10 Ω

R₅ , R₆ are connected in parallel ,

:\to \sf \dfrac{1}{R_{56}}=\dfrac{1}{R_5}+\dfrac{1}{R_6}:→

R

56

1

=

R

5

1

+

R

6

1

:\to \sf \dfrac{1}{R_{56}}=\dfrac{1}{5}+\dfrac{1}{5}:→

R

56

1

=

5

1

+

5

1

:\to \bf R_{56}=\dfrac{5}{2}\ \Omega:→R

56

=

2

5

Ω

R₇ , R₈ are connected in parallel ,

:\to \sf \dfrac{1}{R_{78}}=\dfrac{1}{R_7}+\dfrac{1}{R_8}:→

R

78

1

=

R

7

1

+

R

8

1

:\to \sf \dfrac{1}{R_{78}}=\dfrac{1}{40}+\dfrac{1}{40}:→

R

78

1

=

40

1

+

40

1

:\to \bf R_{78}=20\ \Omega:→R

78

=20 Ω

Now R₁₂ , R₃₄ are connected in series ,

:\to \sf R_{1234}=R_{12}+R_{34}:→R

1234

=R

12

+R

34

:\to \sf R_{1234}=5+10:→R

1234

=5+10

:\to \bf R_{1234}=15\ \Omega:→R

1234

=15 Ω

R₅₆ , R₇₈ are connected in series ,

:\to \sf R_{5678}=R_{56}+R_{78}:→R

5678

=R

56

+R

78

:\to \sf R_{5678}=\dfrac{5}{2}+20:→R

5678

=

2

5

+20

:\to \bf R_{5678}=\dfrac{45}{2}\ \Omega:→R

5678

=

2

45

Ω

Now R₁₂₃₄ , R₅₆₇₈ are connected in parallel ,

:\to \sf \dfrac{1}{R_{12345678}}=\dfrac{1}{R_{1234}}+\dfrac{1}{R_{5678}}:→

R

12345678

1

=

R

1234

1

+

R

5678

1

:\to \sf \dfrac{1}{R_{eq}}=\dfrac{1}{15}+\dfrac{1}{\frac{45}{2}}:→

R

eq

1

=

15

1

+

2

45

1

:\to \sf \dfrac{1}{R_{eq}}=\dfrac{1}{15}+\dfrac{2}{45}:→

R

eq

1

=

15

1

+

45

2

:\to \sf \dfrac{1}{R_{eq}}=\dfrac{3+2}{45}:→

R

eq

1

=

45

3+2

:\to \sf \dfrac{1}{R_{eq}}=\dfrac{5}{45}:→

R

eq

1

=

45

5

:\to \sf \dfrac{1}{R_{eq}}=\dfrac{1}{9}:→

R

eq

1

=

9

1

:\leadsto \textsf{\textbf{R$_{\text{eq}}$ =9\ $\boldsymbol \Omega$}}\ \; \pink{\bigstar}:⇝R

eq

=9 Ω ★

Total resistance b/w A and B is 9 Ω .