An ideal battery drives a current of 5 A through a resistor. After a second resistor of 10 is connected in parallel with it, the current drawn from the battery increases to 6 A . Determine the resistance of the first resistor.

Step-by-step solution

Let the voltage of the ideal battery be V and the resistance of the first resistor be R . Initially, the current is 5 A , so: I₁ = V R = 5 V = 5R When a 10 resistor is connected in parallel, the equivalent resistance becomes: R_ eq = 10R 10 + R The new current is 6 A , so: I₂ = V R_ eq = 6 Substituting V = 5R and R_ eq into the equation: 5R 10R 10 + R = 6 5(10 + R) 10 = 6 10 + R 2 = 6 10 + R = 12 R = 2