In the circuit below, vs = 0 Volts, R1 = 8Ω, R2 = 4Ω. Find i in Amps. Give your answer to 3 decimal places.

Question

BlackTom AI · Accepted Answer

KVL: the algebraic sum of voltages around the loop equals zero, so the resistor drops must sum to the source voltage. Write the loop equation and solve.

1) Relevant concepts
- Ohm’s law: v = iR (voltage across a resistor equals current times resistance).
- Kirchhoff’s Voltage Law (KVL): sum of voltage rises and drops around any closed loop is zero.
- An ideal voltage source with value 0 V provides no net emf (Vs = 0) so it contributes 0 V in the KVL sum.

2) Step-by-step reasoning
Follow the single closed loop shown. The only voltage terms are the drops across R1 and R2 and the source Vs. Apply KVL (going around the loop):
+v_s - v_R1 - v_R2 = 0

Using Ohm’s law for the resistor drops:
v_R1 = i R1, v_R2 = i R2

Substitute into KVL:
v_s - i R1 - i R2 = 0

Factor i:
v_s - i (R1 + R2) = 0

3) Intermediate calculation with given values
Given v_s = 0 V, R1 = 8 Ω, R2 = 4 Ω:
0 - i (8 + 4) = 0
-i (12) = 0

Divide both sides by 12:
i = 0 / 12 = 0 A

4) Why the correct answer is correct
Because the only driving source in the circuit has zero volts, there is no net emf to drive current through the series resistances. The KVL equation therefore forces the loop current to be zero regardless of the resistor values.

5) Why other answers would be incorrect
Any nonzero value would imply a net voltage drop around the loop without a source to supply that voltage. Writing the general formula i = Vs / (R1 + R2) makes this clear: with Vs = 0 the numerator is zero, so i must be zero. There is no division-by-zero or infinities here because the resistances are finite and in series.

Conclusion: The current i = 0 A, which matches the provided answer 0.000.

In the circuit below, vs = 0 Volts, R1 = 8Ω, R2 = 4Ω. Find i in Amps. Give your answer to 3 decimal places.Numerical

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