In the circuit below, vs = -2 Volts, R1 = 9Ω, R2 = 6Ω. Find i in Amps. Give your answer to 3 decimal places.

Question

BlackTom AI · Accepted Answer

Ohm’s law for a single loop with series resistors gives i = Vs / (R1 + R2).

Relevant concepts:
- Series resistances add: R_total = R1 + R2.
- KVL around the loop: the source voltage (when traversed from negative to positive terminal) minus the sum of resistor drops equals zero: +Vs - i(R1+R2) = 0.
- Solving for i yields i = Vs/(R1+R2). A negative value means the actual current flows opposite to the arrow shown.

Step-by-step reasoning and calculation:
1. Add the resistances: R_total = R1 + R2 = 9 Ω + 6 Ω = 15 Ω.
2. Apply KVL as described: +Vs - i·R_total = 0 → i = Vs / R_total.
3. Substitute Vs = -2 V and R_total = 15 Ω:
   i = -2 V / 15 Ω.
4. Compute the quotient: -2 / 15 = -0.133333... A.
5. Round to three decimal places: i = -0.133 A.

Why this answer is correct:
- The formula i = Vs/(R1+R2) directly follows from KVL and Ohm’s law for series circuits. Using the given negative source voltage produces a negative current relative to the arrow direction, meaning the actual current flows opposite the arrow. Numerical division yields -0.133 A when rounded to three decimals.

Why other numeric guesses would be incorrect:
- Any positive 0.133 would ignore the negative sign of Vs (wrong direction).
- Different magnitudes (for example 0.200 A) would come from incorrect addition or division (e.g., dividing by 10 Ω instead of 15 Ω).

Confirming final answer: i = -0.133 A (three decimal places).

In the circuit below, vs = -2 Volts, R1 = 9Ω, R2 = 6Ω. Find i in Amps. Give your answer to 3 decimal places.数值题

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