Match each electrical operational scenario on the left with its corresponding effect on el… — JAMB UTME

Match each electrical operational scenario on the left with its corresponding effect on electrical energy or power on the right.

Doubling the electric current passing through a resistor of constant resistance for a fixed durationIncreases the total heat energy generated by a factor of 4
Halving the operating potential difference applied across a resistor of constant resistanceReduces the rate of electrical energy consumption to one-quarter of its initial value
Connecting two identical resistors in parallel across a constant voltage sourceDoubles the total electrical power drawn from the source compared to a single resistor
Operating a $500\,\text{W}$ appliance continuously for a duration of $2\,\text{hours}$ Results in a total electrical energy consumption of $1.0\,\text{kWh}$

Doubling current quadruples heat energy (

H \propto I^2

); halving voltage reduces power to one-quarter (

P \propto V^2

); connecting two identical resistors in parallel doubles total power (

P \propto 1/R_{\text{eq}}

); and operating a 500 W appliance for 2 hours consumes 1.0 kWh.

Each left-side scenario correctly aligns with fundamental laws of electrical energy and power: doubling current quadruples heat energy via

H \propto I^2

; halving voltage reduces power to one-quarter via

P \propto V^2

; parallel resistor connection halves total resistance and doubles power; and operating a 0.5 kW appliance for 2 hours consumes 1.0 kWh.

Apply Joule's Law of Heating (

H = I^2 R t

) to determine the effect of changing current.

Doubling current from

I

2I

results in

H' = (2I)^2 R t = 4 I^2 R t = 4H

, so heat quadruples.

Heat generation depends on the square of current when resistance and time are constant.

Apply the voltage-power relationship (

P = \frac{V^2}{R}

) for changing voltage.

Halving potential difference to

V/2

yields

P' = \frac{(V/2)^2}{R} = \frac{V^2}{4R} = \frac{P}{4}

, so power drops to one-quarter.

Power is directly proportional to the square of potential difference across a constant resistance.

Calculate parallel resistance and total circuit power.

Equivalent resistance of two parallel resistors is

R/2

. Total power is

P_{\text{total}} = \frac{V^2}{R/2} = 2 \frac{V^2}{R}

, which is double the single resistor power.

Parallel combination reduces net resistance by half, thereby doubling total current and power drawn from a constant voltage source.

Calculate energy consumption in commercial units (kWh).

E = 0.5\,\text{kW} \times 2\,\text{h} = 1.0\,\text{kWh}

Commercial energy is computed by expressing power in kW and time in hours.

Mathematical relationships governing electrical power (

P = I^2 R = \frac{V^2}{R}

), Joule heating (

H = I^2 R t

), and commercial energy units (kWh).

Question