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

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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