A 9.1 Ω resistor, 21.7 Ω resistor, and 22.1 Ω resistor are connected in series across a 121.3 V supply. Calculate the total current in the circuit

Question

A 9.1 Ω  resistor, 21.7 Ω resistor, and 22.1 Ω resistor are connected in series across a 121.3 V supply. Calculate the total current in the circuit

BlackTom AI · Accepted Answer

Restating the problem: You have three resistors in series—9.1 Ω, 21.7 Ω, and 22.1 Ω—across a 121.3 V supply, and you need the total current in the circuit.

Step 1: Add the resistances for series connection. In a series circuit, resistances simply add up. 9.1 + 21.7 + 22.1 = 52.9 Ω.

Step 2: Use Ohm's law to find the current. I = V / R, so I = 121.3 V / 52.9 Ω.

Step 3: Compute the division. 121.3 ÷ 52.9 ≈ 2.293 A (approximately 2.29 A when rounded to two decimal places).

Step 4: Compare with the provided answer. The given answer is 2.25 A. The calculated value is about 2.29 A, which is very close but not exactly 2.25 A. The discrepancy could arise from rounding at intermediate steps or from using a slightly different total resistance (for example, if rounding 52.9 Ω to 53 Ω for a quick estimate, I ≈ 121.3 / 53 ≈ 2.29 A, still near 2.29 A). Depending on the required significant figures, 2.29 A or 2.3 A would be a more precise representation.

Summary note: The exact calculation yields approximately 2.29 A. If you must report a value with two significant figures and accept a minor rounding difference, 2.3 A is a reasonable presentation. The explicit computed current from the given data is about 2.29 A, not exactly 2.25 A.

Physics 1 Hon - Bowles - p06/p07 CFE Practice - Electricity

A 9.1 Ω resistor, 21.7 Ω resistor, and 22.1 Ω resistor are connected in series across a 121.3 V supply. Calculate the total current in the circuit

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