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

To approach this problem, I’ll start by identifying that the three resistors are in series, so their resistances simply add up.

First, add the series resistances: 9.1 Ω + 21.7 Ω + 22.1 Ω = 52.9 Ω.

Next, use Ohm's law for a series circuit to find the current: I = V / R, where V = 121.3 V and R = 52.9 Ω.

Compute the current: I = 121.3 V / 52.9 Ω ≈ 2.293 A. This can be rounded to about 2.29 A (depending on the number of significant figures kept).

Now compare to the provided answer of 2.25 A. The value 2.25 A would imply a slightly larger total resistance (since I = V/R, a smaller current means a larger R). Specifically, 121.3 V divided by 2.25 A equals roughly 53.96 Ω, which does not match the calculated total resistance of 52.9 Ω. This difference is likely due to rounding at intermediate steps or a less precise rounding convention.

In summary, the calculated current using the exact summed resistance is approximately 2.29 A, which is a reasonable representation given standard significant figures. The 2.25 A value appears as a rounded or slightly imprecise alternative.

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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