According to Ohm's law, if the potential difference across a conductor doubles while its resistance stays constant, what happens to the current through it?

It doubles. At fixed resistance, I = V/R. Doubling voltage therefore doubles current. Four times would require a different change. An unchanged current would require resistance to double too. Halving current would require a lower voltage or higher resistance.

What does a potential difference of 6 V mean?

The energy-change magnitude is six joules per coulomb. One volt equals one joule per coulomb. Six volts therefore describes six joules of energy change per coulomb. Coulombs per second measure current. Ohms measure resistance. Power also depends on current, so voltage alone does not determine watts.

How should meters be connected to measure branch current and potential difference across a resistor?

Ammeter in series and voltmeter in parallel. The ammeter carries the branch current through a series connection. The voltmeter spans the resistor's two ends in parallel. Connecting both in series misplaces the voltmeter. Connecting both in parallel misplaces the ammeter. Reversing the correct connections also gives the wrong arrangement.

A uniform wire doubles in length at unchanged volume and resistivity. What happens to its resistance?

It becomes four times the original. Constant volume means that doubling length halves cross-sectional area. In R = ρl/A, both changes increase resistance. The combined factor is 2 ÷ 0.5 = 4. Doubling alone ignores the area change. Unchanged or halved resistance contradicts these stated conditions.

Two resistors, 2 Ω and 4 Ω, are connected in series across 12 V. What current passes through each?

2 A. Series resistance is 2 + 4 = 6 Ω. Current is 12/6 = 2 A through both resistors. Three and six amperes would be the individual branch currents across 12 V in parallel. Nine amperes is their parallel total, rather than the series current.

A constant 2 A flows through a 5 Ω resistor for 10 s. How much thermal energy is produced?

200 J. Joule heat is H = I²Rt. Substitution gives 2² × 5 × 10 = 200 J. Twenty joules omits the full time factor. One hundred joules uses I instead of I². Four hundred joules is twice the calculated energy.

A load draws a constant 1 kW for two hours. How much electrical energy does it use?

2 kWh. Energy equals power multiplied by time. The load uses 1 × 2 = 2 kWh, equivalent to 7.2 × 10⁶ J. Two kilowatts measures power. Half a kilowatt-hour divides instead of multiplying. The listed joule value equals only one kilowatt-hour.

Two resistors, 2 Ω and 4 Ω, are connected in parallel. What is their equivalent resistance?

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Two resistors, 2 Ω and 4 Ω, are connected in parallel. What is their equivalent resistance?

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