Question

five resistances are connected as shown in fig.the equivalent resistance between point A and B is​

Answer 1

Answer:

at each and every step of solving I had simplified the complex figure of arrangement of resistors

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Answer 2

The given connection can also be re-drawn as in 'image 1'

Now, by applying the concept of 'resistors in series' for the resistors $(1,2) \& (4,5)$

we get

$R_{12}=3+7 = 10 ohms\\R_{45} = 10+5 = 15 ohms$

refer to the 'image 2' for the modified circuit

Now, applying the concept of 'resistors in parallel' for $R_{12}, R_3, R_{45}$ we get

$\frac{1}{R_{total}}=\frac{1}{R_{12}} +\frac{1}{R_{3}} +\frac{1}{R_{45}} \\\frac{1}{R_{total}} =\frac{1}{10}+\frac{1}{10}+\frac{1}{15}\\ \frac{1}{R_{total}}=0.1+0.1+0.067\\ \frac{1}{R_{total}}=0.267$

∴$R_{total} = 3.745 ohms$

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