First slide

combination of resistance

Question

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

Easy

A

$\frac{9 V}{35}$
B

$\frac{5 V}{18}$
C

$\frac{5 V}{9}$
D

$\frac{18 V}{5}$

Solution

The circuit given resembles the balanced Wheatstone bridge as $\frac{4}{6} = \frac{2}{3}$
Thus, middle anm 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
$R = \frac{R^{'} \times R^{''}}{R^{'} + R^{''}} = \frac{6 \times 9}{6 + 9} = \frac{54}{15} = \frac{18}{5} Ω$
Hence, current $i = \frac{V}{R} = \frac{V}{18 / 5} = \frac{5 V}{18}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App