The heat generated in a resistor in a period of time t is given as

H = I²Rt

Where, I is the current through the resistor

R is the resistance of the resistor

We know that,

Current, I = V/R where V is the voltage across the resistor (Ohm’s Law)

Case (i): The equivalent resistance, R_eq = 2Ω

So, I = 12/2 = 6A

Hence, H = 6²×2×t = 72t J

Case (ii): The equivalent resistance, R_eq = 4Ω (Series combination)

So, I = 12/4 = 3A

Hence, H = 3²×4×t = 36t J

Case (iii): The equivalent resistance, R_eq = 1Ω (Parallel combination)

So, I = 12/1 = 12A

Hence, H = 12²×1×t = 144t J

Thus, the heat dissipation is maximum for case (iii).

NOTE: For series combination, R_eq=R₁+R₂+…..+R_n (For n resistors)

For parallel combination, (1/R_eq) = (1/R₁) + (1/R₂) +…… + (1/R_n) (For n resistors)