1.

Find the current I flowing through the electric circuit shown in the adjoining figure3.6 2682BAsAs12 3124.5 V2​

Answer»

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


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