First slide

combination of resistance

Question

For the network shown in fig, the value of n current is

Moderate

A

5 V/9
B

9V/35
C

18 V/5
D

5v/18

Solution

The circuit is a balanced Wheatstone bridge. Therefore, the resistance 4 $Ω$ is and can be eliminated. The resistances 4 $Ω$ and 2 $Ω$ are in series and their equivalent resistance is 6 $Ω$ . Similarly, the equivalent resistance of 6 $Ω$ and 3 $Ω$ resistances is 9 $Ω$ The equivalent resistance of the circuit is given by
$\begin{array}{l} R_{eq} = \frac{6 \times 9}{6 + 9} = \frac{18}{5} \\ ∴ i = \frac{V}{R_{eq}} = \frac{V}{(18 / 5)} = \frac{5 V}{18} \end{array}$

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