Question

In R-L circuit R=4 ohms, L= 0.5 H emf of cell =6 volts . work done in changing current from 0.80 A to 0.81 A is ​

Answer 1

Work done in changing current from 0.80 A to 0.81 A is 3.96 × 10⁻⁴ J

Given-

• R = 4 ohms
• L = 0.5 H
• emf = 6 volts

We know that current in the circuit is given as

i = i₀ (1 - e^{-Rt/L})

we know that

i₀ = V/R = 6/4 = 1.5 A

work done is given as  

W = ∫i²RT

W = i₀²∫(1 - e^{-Rt/L}) ² Rdt

W = i₀²R (t - L/2R e^-2Rt/L + 2L/R e^-Rt/L)

when current is 0.80

0.80 = 1.5(1 - e^{-Rt/L})

e^{-Rt/L} = 0.467

t = 0.095

when current is 0.81

e^{-Rt/L} = 0.46

t = 0.097 s

so we have

W = (1.5)²(4)(0.097-0.095)-0.5/8 (-6.489×10⁻³) + 1/4 (0.46-0.467)

W = 3.96 × 10⁻⁴ J

https://brainly.in/question/13908170