Question

Find the current I flowing through the electric circuit shown in the adjoining figure
3.6 2
682
B
As
As
12 312
4.5 V
2​

Answer 1

Explanation:

Since, resistors 1Ω and 3Ω are in series,

∴R

eq1

=1+3=4Ω

Now, resistors 4Ω and 6Ω are in parallel,

∴R

eq2

=4Ω∥6Ω=

4+6

4×6

=

10

24

=2.4Ω

Now, 3.6Ω,2.4Ω and 3Ω are again series,

∴R

eq

=3.6+2.4+3=9Ω

Now, the current flowing the electric current I

I=

R

eq

V

⇒I=

9

4.9

=0.5Amp

⇒I=0.5Amp

solution