Current Electricity - Result Question 104
109. For the network shown in the Fig. the value of the current $i$ is
[2005]
(a) $\frac{9 V}{35}$
(b) $\frac{18 V}{5}$
(c) $\frac{5 V}{9}$
(d) $\frac{5 V}{18}$
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Answer:
Correct Answer: 109. (d)
Solution:
- (d) It is a balanced Wheatstone bridge. Hence resistance $4 \Omega$ can be eliminated.
$\therefore R _{\text{eq }}=\frac{6 \times 9}{6+9}=\frac{18}{5}$
$\therefore i=\frac{V}{R _{eq}}=\frac{5 V}{18}$
