Question

resistance of each 10ohm are connected as shown in the figure the effective resistance between A and G is​

Answer 1

In the above attachment R₁ and R₂ are connected in parallel combination,

we know that,

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.,

1/R = 1/R₁ + 1/R₂

By substituting the values in the formula,

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

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

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

$\dashrightarrow \sf \dfrac{1}{R_{12}} = \dfrac{1}{5}$

$\dashrightarrow \sf R_{12} =5 \: ohms$

Now, R₁₂ and R₃ will be connected in series combination

We know that,

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

R = R₁ + R₂

By substituting the values in the formula,

$\dashrightarrow \sf R_{123} = R_{12} + R_{3}$

$\dashrightarrow \sf R_{123} = 5 + 10$

$\dashrightarrow \sf R_{123} = 15 \: ohms$

Now, R₁₂₃ and R₄ are connected in parallel combination so,

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

$\dashrightarrow \sf \dfrac{1}{R_{1234}} = \dfrac{1}{15} + \dfrac{1}{10}$

$\dashrightarrow \sf \dfrac{1}{R_{1234}} = \dfrac{2 + 3}{30}$

$\dashrightarrow \sf \dfrac{1}{R_{1234}} = \dfrac{5}{30}$

$\dashrightarrow \sf \dfrac{1}{R_{1234}} = \dfrac{1}{6}$

$\dashrightarrow \sf R_{1234} =6 \: ohms$

Now, R₁₂₃₄ and R₅ will be connected in series combination so,

$\dashrightarrow \sf R_{eq} = R_{1234} + R_{5}$

$\dashrightarrow \sf R_{eq} = 6 + 10$

$\dashrightarrow \sf R_{eq} = 16 \: ohms$

Thus, the effective resistance between A and G is 16 ohms.