A uniform metallic wire having resistance 4 Ω is bent to form a square loop (ABCD). 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 amount of current (I) is: ____.

Each side of the square loop is made from the uniform wire whose total resistance is 4 Ω, so the resistance of each side is

$R\_\{\backslash text\{side\}\}\; =\; \backslash frac\{4\backslash \; \backslash Omega\}\{4\}=1\backslash \; \backslash Omega$

The circuit therefore has resistances:

AB = BC = CD = DA = 1 Ω and the diagonal resistor BD = 2 Ω.

The 2 V battery is connected across the opposite corners A and C. We assign the node potentials

$V\_A\; =\; 2\backslash \; \backslash text\{V\},\backslash qquad\; V\_C\; =\; 0\backslash \; \backslash text\{V\}$

and let the unknown potentials at B and D be \(V_B\) and \(V_D\).

Apply Kirchhoff’s current law at node B (sum of currents leaving B is zero):

$(V\_B\; -\; V\_A)/1\; +\; (V\_B\; -\; V\_C)/1\; +\; (V\_B\; -\; V\_D)/2\; =\; 0$

Multiplying by 2 gives

$2(V\_B\; -\; 2)\; +\; 2V\_B\; +\; (V\_B\; -\; V\_D)\; =\; 0\; \backslash ;\backslash Longrightarrow\backslash ;\; 5V\_B\; -\; V\_D\; =\; 4\; \backslash qquad\; (1)$

Apply KCL at node D:

$(V\_D\; -\; V\_A)/1\; +\; (V\_D\; -\; V\_C)/1\; +\; (V\_D\; -\; V\_B)/2\; =\; 0$

Multiplying by 2 gives

$2(V\_D\; -\; 2)\; +\; 2V\_D\; -\; (V\_B\; -\; V\_D)\; =\; 0\; \backslash ;\backslash Longrightarrow\backslash ;\; 5V\_D\; -\; V\_B\; =\; 4\; \backslash qquad\; (2)$

Solving equations (1) and (2) simultaneously:

From (1) \(V_D = 5V_B - 4\).

Substitute into (2):

$5(5V\_B\; -\; 4)\; -\; V\_B\; =\; 4\; \backslash ;\backslash Longrightarrow\backslash ;\; 25V\_B\; -\; 20\; -\; V\_B\; =\; 4\; \backslash ;\backslash Longrightarrow\backslash ;\; 24V\_B\; =\; 24\; \backslash ;\backslash Longrightarrow\backslash ;\; V\_B\; =\; 1\backslash \; \backslash text\{V\}$

Then

$V\_D\; =\; 5(1)\; -\; 4\; =\; 1\backslash \; \backslash text\{V\}$

Thus the potentials at B and D are both 1 V.

The total current supplied by the battery is the sum of the currents on the two branches that leave node A:

$I\; =\; \backslash frac\{V\_A\; -\; V\_B\}\{1\}\; +\; \backslash frac\{V\_A\; -\; V\_D\}\{1\}\; =\; (2\; -\; 1)\; +\; (2\; -\; 1)\; =\; 1\; +\; 1\; =\; 2\backslash \; \backslash text\{A\}$

Therefore, the amount of current flowing through the circuit is

$2\backslash \; \backslash text\{A\}$