1.

Find the current drawn from the battery by network of four resistors shown in the figure?

Answer»

Given: 

V = 3V 

R1 = R2 = R3 = R4 = 10Ω 

Here, R1 and R2 are connected in series 

And, R3 and R4 are connected in series.

∴ \(\frac{1}{R_{eq}}=\frac{1}{R_1\,+\,R_2}+\frac{1}{R_3\,+\,R_4}\)

\(\frac{1}{R_{eq}}=\frac{1}{R_1\,+\,R_2}+\frac{1}{R_3\,+\,R_4}=\frac{1}{20}+\frac{1}{20}=\frac{2}{20}\)

Req = \(\frac{20}{2}\)

Now, by Ohm’s Law

V = I x R

I = \(\frac{V}{R}\) = \(\frac{3}{10}\)

∴ I = 0.3A

Thus, Current drawn by the resistor from the battery is 0.3 A


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