If the resistance of a conductor is doubled while keeping the potential difference across its ends constant, what will happen to the current flowing through it?

Correct — B. Use Ohm's law, V = IR, rearranged as I = V⁄R.

Before: I = V⁄R
Double R, keep V: I′ = V⁄(2R)
Compare: I′ = ½ × V⁄R = I⁄2

The current falls to half → option (b).

Why the others are wrong

• (a)
  It will double — Double is what happens if the potential difference doubles at fixed resistance. Here R went up with V fixed, so I = V⁄R must go down.
• (c)
  It will remain unchanged — Unchanged would need V to double along with R. Only R changed, so I = V⁄R changes with it.
• (d)
  It will quadruple — Quadruple has no route here. Current depends on R to the first power, not squared, and R went up, so the current can only fall.

Ohm's law: the current through a metallic conductor is directly proportional to the potential difference across its ends, provided its temperature stays the same. V = IR, where R is the resistance.

Rearranged, I = V⁄R: at fixed V, current is inversely proportional to resistance. Double R and the current halves. Halve R and it doubles.

Resistance depends on the conductor: R = ρl⁄A, so it grows with length and falls as the cross-section widens.

Key facts

• Ohm's law: V = IR, for a metallic conductor at constant temperature.
• The SI unit of resistance is the ohm (Ω), where 1 Ω = 1 V⁄1 A.
• Resistance R = ρl⁄A, where ρ is the resistivity, measured in ohm metre (Ω m).
• Electric power P = VI = I²R = V²⁄R.

Common traps

• Thinking current rises with resistance because both sit in V = IR: at fixed V, I and R move in opposite directions.
• Ignoring what is held fixed: if the current were held constant instead, doubling R would double the potential difference.

No directly related past PYQ was found.