1.

Find the value of current I in the circuit given below:

Answer»

Resistors across AC and DE are in parallel i.e. RAC and RDE are in parallel.

So

\(\frac1{R'_p}=\frac1{R_{AC}}+\frac1{R_{DE}}\)

⇒ \(\frac1{R'_p}=\frac1{30}+\frac1{30}\)

⇒ \(\frac1{R'_p}=\frac{1+1}{30}+\frac2{30}\)

⇒ R'p = \(\frac{30}2\) = 15Ω

Now R’p and RBC are I series. 

R’s = R’p + RBC 

⇒ R’s = 15 + 15 = 30Ω. 

Again, RAB and R’s are in parallel.

\(\frac{1}{R^"_p}=\frac{1}{R_{AB}}+\frac{1}{R'_s}\)

\(\frac{1}{R^"_p}=\frac{1}{15}+\frac{1}{30}\)

⇒ \(\frac{1}{R^"_p}=\frac{1+2}{30}+\frac{3}{30}\)

⇒ R"p = \(\frac{30}3\) = 10Ω

Equivalent resistance of the above circuit is 10Ω. 

Now, current flowing through the circuit is

l = \(\frac{V}{R} =\frac3{10}=0.3\)

The value of current is 0.3A


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