Question

find ,in the given network of resistors the equivalent resistance between the points A and B, between A and D and between A and C

Answer 1

between Aand B
the equivalent resistance is
$\frac{1}{req} =( \frac{1}{10 + 5} ) + \frac{1}{10}$
(AC and CB in series and then AC+CB and AB in parallel)
1/Req= 1/15+1/10
$\frac{1}{req } = \frac{2 + 3}{30}$
$\frac{1}{req} = \frac{5}{30}$
therefore req between Aand B is 30/5=6 ohm
2.between A and D
$\frac{1}{req} = \frac{1}{10 + 7} + \frac{1}{3}$
$= \frac{1}{17} + \frac{1}{3}$
$= \frac{17 + 3}{51}$
$= \frac{20}{51}$
req between A andD is 51/20=2.55ohm

3. between A and C
equivalent resistance will be either 6 or 5ohm .

solution is same like other

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Answer 2

Answer:

Explanation:

AB= 5 ohm

AC= 30/8 ohm