In the given figure, the value of Vₒ will be _____________ V.

Solution

Physics sir · 4 steps

Nothing leaves at Vₒ, so the three branch currents must add up to zero at that node.

1. Read each branch. In every branch the long plate (positive terminal) faces the resistor and the short plate sits on the grounded rail. So the point between a battery and its resistor is at +E, which is 2 V, 4 V or 6 V.
2. Current at the node. The current from a branch into the Vₒ node is (E - Vₒ)/(1 kΩ). The output terminal is open and draws no current, so by Kirchhoff's current law these three currents sum to zero.

   (2 - Vₒ)/1 + (4 - Vₒ)/1 + (6 - Vₒ)/1 = 0

3. Solve. This gives 12 - 3Vₒ = 0, so Vₒ = 4 V. Check: 1 mA flows out of the 6 V branch, none through the 4 V branch, and 1 mA into the 2 V branch, which balances.
4. So. Vₒ = 4 V.

The common slip: 12. Adding the EMFs as if the batteries were in series gives 12. They are in parallel branches, so the voltages are averaged, not added.

The idea to keep. Batteries with resistors all joined to one open node: write KCL at that node. With equal resistors, the node voltage is just the average of the EMFs.

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