CBSE Class 12 · 2009 · All India · Current ElectricityShort Answer

Calculate the current drawn from the battery in the given network.

Answer: $I = 2 A$ .

Solution

The network is a balanced Wheatstone bridge: the ratio of resistances in the two arms is equal (

\frac{R _{1}}{R _{4}} = \frac{R _{5}}{R _{3}} = \frac{1}{2}

), so no current flows through the bridge (galvanometer) arm

R_{2}

. The two series arms are

R_{1} + R_{5} = 1 + 2 = 3 Ω

and

R_{4} + R_{3} = 2 + 4 = 6 Ω

; these are in parallel:

\frac{1}{R} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \Rightarrow R = 2 Ω

. Hence

I = \frac{V}{R} = \frac{4}{2} = 2 A .

CBSE Class 12 Board · 2009 · All India · Subjective