A resistor develops 300 J of thermal energy in 15 s, when a current of 2 A is passed through it. If the current increases to 3 A, the energy developed in 10 s is ______ J.
To determine the energy developed by a resistor when the current is increased, we start by analyzing the relationship between power, energy, and time. The power P dissipated in a resistor is given by the formula: P = I²R, where I is the current and R is the resistance. The energy developed E over a time period t is then E = P × t = I²Rt. Initially, we have:
Energy (E₁) = 300 J
Time (t₁) = 15 s
Current (I₁) = 2 A
Substituting these into the equation:
E₁ = I₁²R × t₁ = (2 A)² × R × 15 s = 300 J 4R × 15 = 300 R = 5 Ω Now, the current is increased to I₂ = 3 A, and we need to calculate the energy E₂ developed in t₂ = 10 s: E₂ = I₂²Rt₂ = (3 A)² × 5 Ω × 10 s = 9 × 5 × 10 = 450 J Thus, the energy developed in 10 seconds with a current of 3 A is 450 J. This value fits within the specified range (450,450), confirming its accuracy.
