Calculate Current Through Resistor in Circuit Network

To find the current $I$ flowing through the resistor $R = 2\,\Omega$, we analyze the given circuit by identifying and simplifying two nested Wheatstone bridge networks.

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Step 1: Analysis of the First Wheatstone Bridge (Upper Right Sub-circuit)

Let us designate the following nodes in the upper-right sub-circuit:

• Node $A$: Junction immediately after resistor $R$.
• Node $C$: Top-right corner node.
• Node $M$: Central junction between the $2\,\Omega$, $8\,\Omega$, and $4\,\Omega$ resistors.
• Node $E$: Junction connecting the upper bridge, the $10\,\Omega$ resistor, and the lower-right $4\,\Omega$ resistor.

The upper-right network between Node $A$ and Node $E$ consists of:

• $R_{AC} = 1\,\Omega$
• $R_{CE} = 2\,\Omega$
• $R_{AM} = 2\,\Omega$
• $R_{ME} = 4\,\Omega$
• $R_{MC} = 8\,\Omega$ (connecting Nodes $M$ and $C$)

Checking the ratio of resistances for the bridge: $\frac{R_{AC}}{R_{CE}} = \frac{1\,\Omega}{2\,\Omega} = \frac{1}{2}$ $\frac{R_{AM}}{R_{ME}} = \frac{2\,\Omega}{4\,\Omega} = \frac{1}{2}$

Since $\frac{R_{AC}}{R_{CE}} = \frac{R_{AM}}{R_{ME}}$, the bridge is balanced. No current flows through the central branch containing $R_{MC} = 8\,\Omega$, so this resistor can be removed.

The equivalent resistance between Node $A$ and Node $E$ ($R_{AE}$) is calculated as: $R_{AE} = (R_{AC} + R_{CE}) \parallel (R_{AM} + R_{ME})$ $R_{AE} = (1 + 2)\,\Omega \parallel (2 + 4)\,\Omega = 3\,\Omega \parallel 6\,\Omega$ $R_{AE} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\,\Omega$

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Step 2: Analysis of the Second Wheatstone Bridge (Main Network)

Now consider the remaining circuit between Node $A$ and the ground reference line (Node $G$, $0\text{ V}$):

• Node $B$: Junction between the $6\,\Omega$ and $12\,\Omega$ resistors.
• Resistor $R_{AB} = 6\,\Omega$
• Resistor $R_{BG} = 12\,\Omega$
• Resistor $R_{AE} = 2\,\Omega$ (simplified from Step 1)
• Resistor $R_{EG} = 4\,\Omega$
• Resistor $R_{BE} = 10\,\Omega$ (connecting Nodes $B$ and $E$)

Checking the ratio of resistances for this second bridge: $\frac{R_{AB}}{R_{BG}} = \frac{6\,\Omega}{12\,\Omega} = \frac{1}{2}$ $\frac{R_{AE}}{R_{EG}} = \frac{2\,\Omega}{4\,\Omega} = \frac{1}{2}$

Since $\frac{R_{AB}}{R_{BG}} = \frac{R_{AE}}{R_{EG}}$, this bridge is also balanced. Consequently, no current flows through the central $10\,\Omega$ resistor, and it can be removed.

me equivalent resistance between Node $A$ and Ground $G$ ($R_{AG}$) is: $R_{AG} = (R_{AB} + R_{BG}) \parallel (R_{AE} + R_{EG})$ $R_{AG} = (6 + 12)\,\Omega \parallel (2 + 4)\,\Omega = 18\,\Omega \parallel 6\,\Omega$ $R_{AG} = \frac{18 \times 6}{18 + 6} = \frac{108}{24} = 4.5\,\Omega$

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Step 3: Calculation of Total Circuit Resistance and Current $I$

The total equivalent resistance $R_{\text{eq}}$ connected across the $6.5\text{ V}$ battery consists of $R = 2\,\Omega$ in series with $R_{AG}$: $R_{\text{eq}} = R + R_{AG} = 2\,\Omega + 4.5\,\Omega = 6.5\,\Omega$

Using Ohm's Law, the current $I$ flowing through resistor $R$ is: $I = \frac{V}{R_{\text{eq}}} = \frac{6.5\text{ V}}{6.5\,\Omega} = 1\text{ A}$