A circuit has a 5 V source in series with a 2 Ω resistor feeding a node that splits to a 4 Ω resistor and a 6 Ω resistor in parallel to ground. Compute the node voltage.

A circuit has a 5 V source in series with a 2 Ω resistor feeding a node that splits to a 4 Ω resistor and a 6 Ω resistor in parallel to ground. Compute the node voltage.

• A. 3.000 V
• B. 1.250 V
• C. 2.726 V
• D. 3.500 V

Answer: (C)

The node voltage is found by treating the 4 Ω and 6 Ω in parallel as a single equivalent resistor and then using a voltage divider with the 2 Ω series resistor. Compute the parallel: Rp = (4 × 6) / (4 + 6) = 24 / 10 = 2.4 Ω. Total resistance = 2 Ω + 2.4 Ω = 4.4 Ω. Current from the source: I = 5 V / 4.4 Ω ≈ 1.136 A. The node voltage is the drop across the parallel load (the same as across Rp): Vnode = I × Rp ≈ 1.136 × 2.4 ≈ 2.727 V. So the node voltage is about 2.726–2.727 V, matching the given result.

The node voltage is found by treating the 4 Ω and 6 Ω in parallel as a single equivalent resistor and then using a voltage divider with the 2 Ω series resistor.

Compute the parallel: Rp = (4 × 6) / (4 + 6) = 24 / 10 = 2.4 Ω.

Total resistance = 2 Ω + 2.4 Ω = 4.4 Ω.

Current from the source: I = 5 V / 4.4 Ω ≈ 1.136 A.

The node voltage is the drop across the parallel load (the same as across Rp): Vnode = I × Rp ≈ 1.136 × 2.4 ≈ 2.727 V.

So the node voltage is about 2.726–2.727 V, matching the given result.