Question

Find the total current l' flowing through the following circuit:​

Answer 1

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.+3=9Ω

Now, the current flowing the electric current I

I=

R

eq

V

⇒I=

9

4.9

=0.5Amp

⇒I=0.5Amp