Question

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

Answer 1

Resistance of the upper branch FCE are in series.

So, equivalent resistance in the branch FCE = R₁ = 10 Ω + 10 Ω = 20 Ω

Resistance of lower branch FDE are in series

So, equivalent resistance in the branch FDE = R₂ = 10 Ω + 10 Ω = 20 Ω

The two resistance R₁ and R₂ are parallel to each other.

If R is the equivalent resistance of network, then

\(\frac 1R = \frac 1{20}+\frac 1{20} \)

\(= \frac 2{20} \)

\(= \frac 1{10}\)

or R = 10 Ω

Current, V = IR

\(I = \frac VR\)

\(= \frac{3V}{10 \Omega}\)

= 0.3 A

Answer 2

Given: 

V = 3V 

R₁ = R₂ = R₃ = R₄ = 10Ω 

Here, R₁ and R₂ are connected in series 

And, R₃ and R₄ are connected in series.

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

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

R_eq = \(\frac{20}{2}\)

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