Using Electricity Safely

Use $I=V/R$ where $V=230$ V and $R=100,000$ Ω.

$I=230/100,000=0.0023$ A

$0.0023\times 1000=2.3$ mA

From the current-effect table, 2.3 mA falls between 1 mA (barely perceptible) and 5 mA (painful shock) — noticeable but probably non-fatal.

230 mA would require a body resistance of only 1,000 Ω ($230/1000=0.23$ A = 230 mA). Dry skin resistance is ~100,000 Ω, so this is far too high.

23 mA would require resistance of about 10,000 Ω, not 100,000 Ω. Eliminate this option.

$0.23$ mA would require $R=230/0.00023\approx 1,000,000$ Ω, far higher than typical body resistance. Eliminate.

2.3 mA is the only value consistent with $I=230/100,000=0.0023$ A = 2.3 mA, and matches the 'noticeable but non-fatal' range from the table.

Dry skin resistance is approximately 100,000 Ω; wet skin resistance drops to approximately 1,000 Ω — a factor of 100 lower.

From $I=V/R$, if $R$ decreases by a factor of 100 and $V$ stays the same (230 V), then $I$ increases by a factor of 100.

Dry: $I=230/100,000=2.3$ mA. Wet: $I=230/1,000=230$ mA. The wet-skin current is in the life-threatening range (>200 mA).

It is the dramatic drop in resistance — not any change in voltage — that causes the dangerous increase in current through the body.

Wet skin has lower resistance, not higher. This option contradicts the notes. Eliminate.

Water does not increase the supply voltage. Voltage is determined by the supply, not the person's skin condition. Eliminate.

This is physiologically incorrect and not supported by the notes. Current through the body still travels the same internal path. Eliminate.

The only remaining option — lower resistance causing greater current — is consistent with Ohm's Law and the study notes.