In the given circuit, the value of | (I₁ + I₃)/I₂ | is _____________

Solution

Physics sir · 4 steps

Fix potentials at the three junctions first; then every current is just Δ V / R.

1. Name the nodes. Call the left wire L, the junction between the two cells M, and the right wire R. Take V_M = 0. The longer plate of a cell is its positive terminal.
2. Use the cells. The 10 V cell has its long plate on the L side, so V_L = +10 V. The 20 V cell has its long plate on the M side, so V_R = -20 V.

   I = (Vₛₜₐᵣₜ - V_(end))/R

3. Read each current. I₁ and I₂ are drawn from M to L, so each equals (0 - 10)/10 = -1 A. I₃ is drawn from L to R through the bottom resistor, so I₃ = (10 - (-20))/10 = 3 A.
4. So. (I₁ + I₃)/I₂ = (-1 + 3)/-1 = -2, so the magnitude is 2.

The common slip: 4. Dropping the minus sign on I₁ and I₂ (taking them as +1 A) gives (1 + 3)/1 = 4, not 2. Keep the signs set by the arrow directions.

The idea to keep. In circuits where cells sit directly between junctions, fix the node potentials first. Then each current follows from Δ V / R, and a negative sign just means it flows opposite to the drawn arrow.

Answer from the NTA final answer key. Solution by PrepWiser.