Q Find the currents in the given circuit

⚡ Find Current in Each Branch

🔍 Given Circuit

Let the branch currents be:

• $I_1$ in the lower branch
• $I$​ in the middle branch
• $I_3$​ in the upper branch

Using Kirchhoff’s Current Law (KCL) at junction F:$$I_1 + I_2 = I_3$$

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🔄 Loop ABCF

Applying Kirchhoff’s Loop Rule (KVL):$$-(I_1+I_2)(1)-2-2I_2-1=0$$Expanding:$$-I_1-I_2-2-2I_2-1=0$$ $$-I_1-3I_2-3=0$$

Therefore,$$I_1+3I_2=-3$$

✅ Equation (1)

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🔄 Loop ABDE

Applying KVL:$$-(I_1+I_2)(1)-2+4-2I_1=0$$

Expanding:$$-I_1-I_2-2+4-2I_1=0$$$$-3I_1-I_2+2=0$$

Therefore,$$3I_1+I_2=2$$

✅ Equation (2)

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✏️ Solving Equations (1) and (2)

From Equation (1):$$I_1+3I_2=-3$$

Multiply by 3:$$3I_1+9I_2=-9$$

Equation (2):$$3I_1+I_2=2$$

Subtracting:$$8I_2=-11$$$$I_2=-\frac{11}{8}A$$

✅ Current in middle branch$$\boxed{I_2=-\frac{11}{8}A}$$​

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Substituting into Equation (2):$$3I_1-\frac{11}{8}=2$$ $$3I_1=\frac{27}{8}$$$$I_1=\frac{9}{8}A$$

✅ Current in lower branch$$\boxed{I_1=\frac{9}{8}A}$$

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🔌 Finding $I_3$

Using KCL:$$I_3=I_1+I_2$$​ $$I_3=\frac{9}{8}-\frac{11}{8}$$​ $$I_3=-\frac{2}{8}$$ $$I_3=-\frac{1}{4}A$$

✅ Current in upper branch$$\boxed{I_3=-\frac{1}{4}A}$$​

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🎯 Final Currents

✅ Upper Branch:$$\boxed{I_3=-\frac{1}{4}A}$$​

✅ Middle Branch:$$\boxed{I_2=-\frac{11}{8}A}$$

✅ Lower Branch:$$\boxed{I_1=\frac{9}{8}A}$$

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⚡ Find Potential Difference Between A and D

Using the upper branch:$$V_A-\left(-\frac{1}{4}\times1\right)-2=V_D$$ $$V_A+\frac{1}{4}-2=V_D$$$$V_A-\frac{7}{4}=V_D$$

Therefore,$$V_D-V_A=-\frac{7}{4}V$$

✅ Potential difference between D and A$$\boxed{V_D-V_A=-\frac{7}{4}V}$$

or$$\boxed{V_A-V_D=\frac{7}{4}V}$$​

🎉 Hence, the currents in all branches and the potential difference between A and D have been determined.