Question

Find the quivalent resistance across the terminals A and B?​

Answer 1

Given :

According to the given figure,

• R₁ = 3 ohms
• R₂ = 3 ohms
• R₃ = 4 ohms
• R₄ = 5 ohms
• R₅ = 2 ohms
• R₆ = 3 ohms
• R₇ = 4 ohms
• R₈ = 3 ohms
• R₉ = 4 ohms

To find :

The quivalent resistance across the terminals A and B.

Solution :

According to the figure,

R₁ and R₂ are connected in parallel combination (Refer fig-1 from the given attachment),

» 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.,

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

$\dashrightarrow \sf \dfrac{1}{R_{12}} = \dfrac{1}{3} + \dfrac{1}{3}$

$\dashrightarrow \sf \dfrac{1}{R_{12}} = \dfrac{2}{3}$

$\dashrightarrow \sf R_{12} = \dfrac{3}{2}$

$\dashrightarrow \sf R_{12} = 1.5 \: ohms$

Now, R₃ and R₁₂ are connected in series (Refer fig - 2 from the attachment),

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

$\dashrightarrow \sf R_{123} = R_{12} + R_3$

$\dashrightarrow \sf R_{123}= 1.5 + 4$

$\dashrightarrow \sf R_{123} = 5.5 \: ohms$

Now, R₁₂₃ and R₄ are connected in parallel combination (Refer fig-3 from the given attachment),

$\dashrightarrow \sf \dfrac{1}{R_{1234}} = \dfrac{1}{R_{123}} + \dfrac{1}{R_4}$

$\dashrightarrow \sf \dfrac{1}{R_{1234}} = \dfrac{1}{5.5} + \dfrac{1}{5}$

$\dashrightarrow \sf \dfrac{1}{R_{1234}} = \dfrac{5 + 5.5}{27.5}$

$\dashrightarrow \sf \dfrac{1}{R_{1234}} = \dfrac{10.5}{27.5}$

$\dashrightarrow \sf R_{1234} = \dfrac{27.5}{10.5}$

$\dashrightarrow \sf R_{1234} = 2.61 \: ohms$

Now, R₁₂₃₄ , R₈ and R₅ ,R₆ are connected series combination (refer fig-4 from the given attachment),

$\dashrightarrow \sf R_{12348} = R_{1234} + R_{8}$

$\dashrightarrow \sf R_{12348} = 2.61 + 3$

$\dashrightarrow \sf R_{12348} = 5.61 \: ohms$

similarly,

$\dashrightarrow \sf R_{56}= R_5 + R_6$

$\dashrightarrow \sf R_{56} = 2 + 3$

$\dashrightarrow \sf R_{56} = 5 \: ohms$

Now, R₁₂₃₄₈ and R₅₆ are connected in parallel combination ( refer fig-5 from the given attachment),

$\dashrightarrow \sf \dfrac{1}{R_{1234856}} = \dfrac{1}{R_{12348}} + \dfrac{1}{R_{56}}$

$\dashrightarrow \sf \dfrac{1}{R_{1234856}} = \dfrac{1}{5.61} + \dfrac{1}{5}$

$\dashrightarrow \sf \dfrac{1}{R_{1234856}} = \dfrac{10.61}{28.5}$

$\dashrightarrow \sf \dfrac{1}{R_{1234856}} = \dfrac{28.05}{10.61}$

$\dashrightarrow \sf \dfrac{1}{R_{1234856}} = 2.65ohms$

Finally R₇, R₁₂₃₄₈₅₆ and R₉ come in series combination,

$\dashrightarrow \sf R_{eq} = R_7 + R_{1234856} + R_9$

$\dashrightarrow \sf R_{eq} = 4 + 2.65 + 4$

$\dashrightarrow \underline{ \boxed{ \sf R_{eq} = 10.65 \: ohms}}$

thus, the equivalent resistance across the terminals A and B is 10.65 ohms.

hence Option (b) 10.65 ohms is correct.