Question

A16 ohm resistance wire is bent to form a square. A source of 9 volt is connected across one of its sides calculate the potential difference across any one of its diagonal

Answer 1

Since wire is bent into form of a square, so each side of square will have $=4ohm$ resistance as shown below.

Here wire AD, DC and CB are in series and AB will be parallel to it,

So, eqivalent resistance will be,$=3R\parallel R⟹R_{eq}=\frac{3R\times R}{3R+R}=\frac{3R}{4}=\frac{3\times 4}{4}=3ohm$

Current in a loop$I=\frac{V}{R_{eq}}=\frac{9V}{3}=3A$

Since AD , DC , CB resistors are in series, they will equally divide the potential difference from battery.

So, potential difference between end A and C= Potential drop in AD+ Potential drop in DC $V_{AC}=6volt$