Magnitude of Current in Symmetry Resistance Circuit with Two Batteries

To determine the magnitude of the current flowing through each resistor, we analyze the circuit using nodal analysis and symmetry.

1. Circuit Symmetry and Node Definitions

Let us define the potentials at key junctions (nodes) in the circuit:

• Let the central junction be node $O$ with potential $V_O = 0\text{ V}$ (reference node).
• Let the leftmost node be $L$ with potential $V_L$.
• Let the rightmost node be $R$ with potential $V_R$.
• Let the top node be $T$ with potential $V_T$.
• Let the bottom node be $B$ with potential $V_B$.

Due to top-bottom symmetry across the horizontal axis passing through $L, O, R$, the potential at node $T$ equals the potential at node $B$: $V_T = V_B$

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2. Applying Kirchhoff's Current Law (KCL)

At Node $T$:

Applying KCL at node $T$: $\frac{V_T - V_L}{R_6} + \frac{V_T - V_R}{R_7} + \frac{V_T - V_O}{R_2} = 0$

Given $R_1 = R_2 = \dots = R_8 = 1\ \Omega$ and $V_O = 0\text{ V}$: $(V_T - V_L) + (V_T - V_R) + V_T = 0$ $3V_T = V_L + V_R \implies V_T = \frac{V_L + V_R}{3}$

Since $V_B = V_T$, we also have: $V_B = \frac{V_L + V_R}{3}$

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At Node $L$:

The branch between $L$ and $O$ contains battery $\mathcal{E}_2 = 6\text{ V}$ and resistor $R_3$. The potential just after the battery (towards $R_3$) is $(V_L - \mathcal{E}_2)$.

Applying KCL at node $L$: $\frac{V_L - V_T}{R_6} + \frac{V_L - V_B}{R_5} + \frac{(V_L - \mathcal{E}_2) - V_O}{R_3} = 0$

Substituting $V_B = V_T$ and $V_O = 0\text{ V}$: $2(V_L - V_T) + V_L - 6 = 0$ $3V_L - 2V_T = 6$

Substituting $V_T = \frac{V_L + V_R}{3}$: $3V_L - 2\left(\frac{V_L + V_R}{3}\right) = 6$ $9V_L - 2V_L - 2V_R = 18$ $7V_L - 2V_R = 18 \quad \text{--- (Equation 1)}$

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At Node $R$:

The branch between $O$ and $R$ contains battery $\mathcal{E}_1 = 12\text{ V}$ and resistor $R_1$. The potential just after the battery (towards $R_1$) is $(V_O - \mathcal{E}_1) = -12\text{ V}$.

Applying KCL at node $R$: $\frac{V_R - V_T}{R_7} + \frac{V_R - V_B}{R_8} + \frac{V_R - (V_O - \mathcal{E}_1)}{R_1} = 0$

Substituting $V_B = V_T$ and $V_O = 0\text{ V}$: $2(V_R - V_T) + V_R - (-12) = 0$ $3V_R - 2V_T = -12$

Substituting $V_T = \frac{V_L + V_R}{3}$: $3V_R - 2\left(\frac{V_L + V_R}{3}\right) = -12$ $9V_R - 2V_L - 2V_R = -36$ $-2V_L + 7V_R = -36 \quad \text{--- (Equation 2)}$

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3. Solving for Node Potentials

From Equation 1: $V_R = \frac{7V_L - 18}{2}$

Substituting $V_R$ into Equation 2: $-2V_L + 7\left(\frac{7V_L - 18}{2}\right) = -36$ $-4V_L + 49V_L - 126 = -72$ $45V_L = 54 \implies V_L = \frac{54}{45} = 1.2\text{ V}$

Now calculating $V_R$: $V_R = \frac{7(1.2) - 18}{2} = \frac{8.4 - 18}{2} = -4.8\text{ V}$

Now calculating $V_T$ and $V_B$: $V_T = V_B = \frac{1.2 - 4.8}{3} = \frac{-3.6}{3} = -1.2\text{ V}$

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4. Calculating Currents Through Resistors

1. Current through $R_1$: $I_{R_1} = \frac{|V_R - (V_O - \mathcal{E}_1)|}{R_1} = \frac{|-4.8 - (-12)|}{1} = 7.2\text{ A}$ (Statement A is correct)

2. Current through $R_2$: $I_{R_2} = \frac{|V_O - V_T|}{R_2} = \frac{|0 - (-1.2)|}{1} = 1.2\text{ A}$ (Statement B is correct)

3. Current through $R_3$: $I_{R_3} = \frac{|V_O - (V_L - \mathcal{E}_2)|}{R_3} = \frac{|0 - (1.2 - 6)|}{1} = 4.8\text{ A}$ (Statement C is correct)

4. Current through $R_5$: $I_{R_5} = \frac{|V_L - V_B|}{R_5} = \frac{|1.2 - (-1.2)|}{1} = 2.4\text{ A}$ (Statement D is correct)

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Conclusion

All four statements (A), (B), (C), and (D) are correct.