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Q.Apply the Kirchhoff's loop rule to the closed loop ABCA of the network shown in the figure, and write the relevant equation.

A kite/bridge network A, B, C, D, E with 4 ohm, 1 ohm and 2 ohm resistors, a 10 V and a 5 V cell, and branch currents I1, I2, I3 — Class 12 Physics Kirchhoff question
Figure
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Kirchhoff's loop (voltage) rule says the algebraic sum of potential changes around any closed loop is zero; applying it to loop ABCA gives an equation relating I1I_1, I2I_2, I3I_3.

Kirchhoff's Voltage Law (loop rule): In any closed loop of a circuit, the algebraic sum of the EMFs equals the algebraic sum of the IR drops, i.e. ∑ε=∑IR\sum \varepsilon = \sum IR, provided a consistent sign convention is used while traversing the loop.

Applying it to loop A → B → C → A (as shown in the figure):

  • Branch A→B carries the 10 V source in series with the 4Ω resistor, current I2I_2 flowing A→B (same sense as our traversal) → contributes +10 V+10\text{ V} (EMF) and a drop of 4I24I_2.
  • Branch B→C is the 2Ω resistor carrying current (I2+I3)(I_2+I_3), also flowing B→C (same sense as traversal) → drop of 2(I2+I3)2(I_2+I_3).
  • Branch C→A is the 1Ω resistor, but the labelled current I1I_1 actually flows A→C — i.e. OPPOSITE to our C→A traversal — so its contribution enters with a negative sign: −I1(1)-I_1(1). …

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