Physics, asked by DEVPRASSATH996, 11 months ago

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

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Answered by aristocles
8

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 3.96\times 10^{-4} J

As we know that current in the circuit is given as

i = i_o(1 - e^{-Rt/L})

here we know that

i_o = \frac{V}{R}

i_o = \frac{6}{4}

i_o = 1.5 A

so we will have work done given as

W = \int i^2 R dt

W = i_o^2\int (1 - e^{-Rt/L})^2 R dt

W = i_o^2 R \int (1 + e^{-2Rt/L} - 2e^{-Rt/L}) dt

W = i_o^2 R(t - \frac{L}{2R}e^{-2Rt/L} + \frac{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) - \frac{0.5}{2(4)}(-6.489 \times 10^{-3}) + \frac{2(0.5)}{4} (0.46 - 0.467)

W = 3.96 \times 10^{-4} J

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Topic : LR charging circuit

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