An electric coil is rated 400W, 200V. It is cut into two equal parts and connected in parallel to the same source of 200V. Calculate the percentage increase in energy produced per second.

Step-by-step solution

Let the original resistance of the coil be R . The original power is P₁ = V^2 R = 400 W. When the coil is cut into two equal parts, the resistance of each part becomes R 2 . When these two parts are connected in parallel, the equivalent resistance R_ eq is given by: R_ eq = R 2 R 2 R 2 + R 2 = R 4 The new power P₂ when connected to the same voltage source V is: P₂ = V^2 R_ eq = V^2 R 4 = 4 V^2 R = 4 P₁ The increase in power is P = P₂ - P₁ = 4 P₁ - P₁ = 3 P₁ . The percentage increase in energy produced per second is