JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 20)
Five equal resistances are connected in a network as shown in figure. The net resistance between the points A and B is :
_26th_February_Morning_Shift_en_20_1.png)
$${{3R} \over 2}$$
$${{R} \over 2}$$
2R
R
Explanation
Given all resistances have same resistance R.
Now, we can redraw the circuit as below
_26th_February_Morning_Shift_en_20_2.png)
Let resistances be R1, R2, R3 and R4.
$$\because$$ $${{{R_1}} \over {{R_3}}} = {{{R_2}} \over {{R_4}}}$$
So, circuit will behave as a Wheatstone bridge and no current will flow through middle resistor.
$$\therefore$$ $${R_{eq}} = {{({R_1} + {R_2})({R_3} + {R_4})} \over {({R_1} + {R_2}) + ({R_3} + {R_4})}}$$
$$ = {{(R + R)(R + R)} \over {(R + R) + (R + R)}}$$
$$ = R$$
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