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

$\frac{9 V}{35}$

$\frac{5 V}{18}$

$\frac{5 V}{9}$

$\frac{18 V}{5}$

Solution

$\frac{5 V}{18}$

The circuit given resembles the balanced Wheatstone Bridge as $\frac{4}{6} = \frac{2}{3}$
Thus, a middle arm containing 4 Ω resistance will be ineffective and no current flows through it. The equivalent circuit is shown as below:

Net resistance of AB and BC

R' = 4 + 2 = 6 Ω

Net resistance of AD and DC

R'' = 6 + 3 = 9 Ω

Thus parallel combination of R' and R'' gives

$R = \frac{R' \times R''}{R' + R''} = \frac{6 \times 9}{6 + 9} = \frac{54}{15} R = \frac{18}{5} Ω Hence current i = \frac{V}{R} = \frac{V}{\frac{18}{5}} i = \frac{5 V}{18}$

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Current Electricity

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

$\frac{9 V}{35}$

$\frac{5 V}{18}$

$\frac{5 V}{9}$

$\frac{18 V}{5}$

Some More Questions From Current Electricity Chapter

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For Daily Free Study Material Join wiredfaculty

Current Electricity

For the network shown in the figure, the value of the current i is 9 V35 5 V18 5 V9 18 V5

Some More Questions From Current Electricity Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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

$\frac{9 V}{35}$

$\frac{5 V}{18}$

$\frac{5 V}{9}$

$\frac{18 V}{5}$