Question

find the Curent flausing the ough
the following electric circuit.
3.6
6
3
ma
is
Amunt
38
#HHE
4.57​

Answer 1

Answer:

see below

Step-by-step 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