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For the network shown in the figure the value of the current $i$ is -
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A
$\dfrac{18V}{5}$
B
$\dfrac{5V}{9}$
C
$\dfrac{9V}{35}$
D
$\dfrac{5V}{18}$
Detailed Solution
The resistances $4\,\Omega$, $6\,\Omega$, $2\,\Omega$ and $3\,\Omega$ form the four arms of a Wheatstone bridge, with the middle $4\,\Omega$ resistor as the galvanometer arm.
Check the balance: $\dfrac{4}{6} = \dfrac{2}{3}$, so the bridge is balanced and no current flows through the middle $4\,\Omega$ resistor; it can be removed.
The circuit then reduces to two parallel branches: $(4 + 2) = 6\,\Omega$ and $(6 + 3) = 9\,\Omega$.
Equivalent resistance $= \dfrac{6 \times 9}{6 + 9} = \dfrac{54}{15} = \dfrac{18}{5}\,\Omega$
Current drawn from the source: $i = \dfrac{V}{18/5}$
$\Rightarrow i = \dfrac{5V}{18}$
Check the balance: $\dfrac{4}{6} = \dfrac{2}{3}$, so the bridge is balanced and no current flows through the middle $4\,\Omega$ resistor; it can be removed.
The circuit then reduces to two parallel branches: $(4 + 2) = 6\,\Omega$ and $(6 + 3) = 9\,\Omega$.
Equivalent resistance $= \dfrac{6 \times 9}{6 + 9} = \dfrac{54}{15} = \dfrac{18}{5}\,\Omega$
Current drawn from the source: $i = \dfrac{V}{18/5}$
$\Rightarrow i = \dfrac{5V}{18}$