A uniform metallic wire having resistance 4 Ω is bent to form a square loop (ABCD) (see figure). A resistance of 2 Ω is connected between points B and D and a battery of 2 V is connected across points A and C as shown in the figure. Now the value of current (l) is: (a) 2 A (b) 8 A (c) 4.5 A (d) 4 A

Answer: (a). Each side of the square has 4/4 = 1 Ω. AB: BC = AD: DC = 1: 1, so the bridge is balanced and the 2 Ω in BD carries no current. Paths ABC and ADC are each 2 Ω, in parallel: R = 1 Ω. I = E/R = 2/1 = 2 A So the answer is (a) 2 A.

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Kirchhoff's Laws & Measuring Instruments

A uniform metallic wire having resistance $4\ \Omega$ is bent to form a square loop (ABCD) (see figure). A resistance of $2\ \Omega$ is connected between points $B$ and $D$ and a battery of 2 V is connected across points $A$ and $C$ as shown in the figure. Now the value of current $(l)$ is:

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Answer: (a)

Each side of the square has $\frac{4}{4} = 1\ \Omega$ .
AB : BC = AD : DC = 1 : 1, so the bridge is balanced and the $2\ \Omega$ in BD carries no current.
Paths ABC and ADC are each $2\ \Omega$ , in parallel: $R = 1\ \Omega$ .
$I = \frac{E}{R} = \frac{2}{1} = 2\text{ A}$
So the answer is (a) 2 A.

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