Find the total resistance between A and B of the surface
R1=R2=10
R3=R4=20
R5=R6=5
R7=R8=40
Answers
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 ,
R₃ , R₄ are connected in parallel ,
R₅ , R₆ are connected in parallel ,
R₇ , R₈ are connected in parallel ,
Now R₁₂ , R₃₄ are connected in series ,
R₅₆ , R₇₈ are connected in series ,
Now R₁₂₃₄ , R₅₆₇₈ are connected in parallel ,
Total resistance b/w A and B is 9 Ω .
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 .