For the network shown in the figure the value of the current i is (a) 9V/35 (b) 18V/5 (c) 5V/9 (d) 5V/18

Answer: (d). Ratios 4/2 = 6/3, so the bridge is balanced and the middle 4 Ω carries no current. Paths: 4 + 2 = 6 Ω and 6 + 3 = 9 Ω in parallel. R = (6 × 9)/(15) = 18/5 Ω i = V/R = 5V/18 So the answer is (d).

For the network shown in the figure the value of the current i is | NEET 2005 (AIPMT)

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Kirchhoff's Laws & Measuring Instruments

For the network shown in the figure the value of the current $i$ is

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Answer: (d)

Ratios $\frac{4}{2} = \frac{6}{3}$ , so the bridge is balanced and the middle $4\ \Omega$ carries no current.
Paths: $4 + 2 = 6\ \Omega$ and $6 + 3 = 9\ \Omega$ in parallel.
$R = \frac{6 \times 9}{15} = \frac{18}{5}\ \Omega$
$i = \frac{V}{R} = \frac{5V}{18}$
So the answer is (d).

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