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Physics
Twelve resistors each of resistance 16 Ω are connected in the circuit as shown. The net resistance between A and B is
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Q. Twelve resistors each of resistance $16\,\Omega$ are connected in the circuit as shown. The net resistance between $A$ and $B$ is
BITSAT
BITSAT 2019
A
$1\,\Omega$
B
$2\,\Omega$
C
$3\,\Omega$
D
$4\,\Omega$
Solution:
The equivalent circuit is shown in figure.
The equivalent resistance of each parallel branch is
$R_{P}=R\|R\| R=(1 / R+1 / R+$ $1 / R)^{-1}=R / 3$
Thus, net resistance is
$R_{\text {net }}=(R / 3+R / 3+R / 3)\|(R / 3)=R\|(R / 3)$
$=\frac{R \times R / 3}{R+R / 3}=R / 4=16 / 4=$
$4\, \Omega$