The resistance between A and B in the given figure will be(in ohm):

Question

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Answer 1

In the given circuit, suppose, $r = 10 Ω$
$r_{1} = 10 Ω$
$r_{2} = 20 Ω$
and, $r_{3} = 50 Ω$
(Please refer the diagram.)
Now, $r_{1}, r_{2}, r_{3}$ are connected in parallel. Suppose, resultant of these three resistances is $R_{e}$ .
Therefore, $\frac{1}{r _{e}} = \frac{1}{r _{1}} + \frac{1}{r _{2}} + \frac{1}{r _{3}} = \frac{1}{1 0} + \frac{1}{2 0} + \frac{1}{5 0} = \frac{1 7}{1 0 0}$
Hence, $r_{e} = \frac{1 0 0}{1 7} Ω$
Now, resistances $r$ and $r_{e}$ are connected in series.
Therefore, equivalent resistance in the given circuit, $R = r + r_{e}$
$R = 10 + \frac{1 0 0}{1 7} = \frac{2 7 0}{1 7} Ω$ , which is more than $10 Ω$ , but less than $20 Ω$ .

Answer 2

In the given circuit, suppose,

r = 10 Ω

r_{1} = 10 Ω

r_{2} = 20 Ω

and,

r_{3} = 50 Ω

(Please refer the diagram.)
Now,

r_{1}, r_{2}, r_{3}

are connected in parallel. Suppose, resultant of these three resistances is

R_{e}

.
Therefore,

\frac{1}{r _{e}} = \frac{1}{r _{1}} + \frac{1}{r _{2}} + \frac{1}{r _{3}} = \frac{1}{1 0} + \frac{1}{2 0} + \frac{1}{5 0} = \frac{1 7}{1 0 0}

Hence,

r_{e} = \frac{1 0 0}{1 7} Ω

Now, resistances

r

and

r_{e}

are connected in series.
Therefore, equivalent resistance in the given circuit,

R = r + r_{e}

R = 10 + \frac{1 0 0}{1 7} = \frac{2 7 0}{1 7} Ω

, which is more than

10 Ω

, but less than

20 Ω

.