Question

Calculate the equivalent resistance of the network across the
points A and B shown in figure.
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Answer 1

Here in this diagram, say:

R1 = 4Ω

R2 = 8Ω  

R3 = 4Ω  

R4 = 8Ω  

When we want to find the resistance across points A and B then resistance R1 and R2 are connected in series and resistances R3 and R4 are also connected in series.

Thus,  R5 = R1 + R2 = 4Ω + 8Ω = 12Ω  And,

R6 = R3 + R4 = 4Ω + 8Ω = 12Ω  

Now, this R5 and R6 are connected to each other in parallel. Thus the equivalent resistance RAB across points A and B is:

1/RAB=1/R5+1/R6

1/RAB=1/12+1/12

1/RAB=2/12+1/6

∴ RAB = 6Ω

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Answer 2

Given :

Four resistors are connected in a circuit

• Resistor ED = 4Ω
• Resistor DC = 8Ω
• Resistor EF = 4Ω
• Resistor FC = 8Ω

To find :

The equivalent resistance of the network across the points A and B.

Solution :

Firstly let's consider Resistor ED as R₁, Resistor DC as R₂, Resistor EF R₃ and Resistor FC as R₄

When we rearrange the given figure we can see that R₁, R₂ and R₃ R₄ are connected in series combination,

» The combined resistance of any number of resistance connected in series is equal to the sum of the individual resistances. i.e.,

$\dashrightarrow \sf R_{12} = R_1 + R_2$

$\dashrightarrow \sf R_{12} =4 + 8$

$\dashrightarrow \sf R_{12} = 12 \: ohms$

similarly,

$\dashrightarrow \sf R_{34} = R_3 + R_4$

$\dashrightarrow \sf R_{34} = 4 + 8$

$\dashrightarrow \sf R_{34} = 12 \: ohms$

Now, Resistor EDC and EFC are connected in parallel combination that is R₁₂ and R₃₄ are connected in parallel combination,

» The reciprocal of the combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistances. .i.e.,

$\dashrightarrow \sf \dfrac{1}{R _{eq} } =\dfrac{1}{R_{12}} + \dfrac{1}{R_{34}}$

$\dashrightarrow \sf \dfrac{1}{R _{eq} } =\dfrac{1}{12} + \dfrac{1}{12}$

$\dashrightarrow \sf \dfrac{1}{R _{eq} } =\dfrac{1 + 1}{12}$

$\dashrightarrow \sf \dfrac{1}{R _{eq} } =\dfrac{2}{12}$

$\dashrightarrow \sf \dfrac{1}{R _{eq} } =\dfrac{1}{6}$

$\dashrightarrow \sf R _{eq} =6 \: ohms$

Thus, the equivalent resistance of the network across the points A and B is 6 ohms.

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