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The initial rate of increase of current when a battery of emf $6 \mathrm{~V}$ is connected in series with an inductance of $2 \mathrm{H}$ and resistance of $12 \Omega$ is
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Verified Answer
The correct answer is:
$3 \mathrm{~A} \mathrm{~s}^{-1}$
Given that, emf of battery, $E=6 \mathrm{~V}$
Resistance, $R=12 \Omega$
Inductance, $L=2 \mathrm{H}$
We know that, the instantaneous current in the circuit containing $R$ and $L$ in series with battery of emf $E$ is
$$
I=I_0\left(1-e^{\frac{-R}{L} t}\right)
$$
where, maximum current at steady state $I_0=E / R$
$$
\therefore \quad I=\frac{E}{R}\left(1-e^{\frac{-R}{L} t}\right)
$$
Now, the rate of increase of current,
$$
\frac{d I}{d t}=\frac{E}{R}\left[0-\left(\frac{-R}{L}\right) e^{\frac{-R}{L} t}\right]=\frac{E}{L} e^{\frac{-R}{L} t}
$$
At $(t=0)$ initial condition,
$$
\left.\frac{d I}{d t}\right|_{t=0}=\frac{E}{L} e^0=\frac{E}{L}=\frac{6}{2}=3 \mathrm{~A} / \mathrm{s}
$$
Resistance, $R=12 \Omega$
Inductance, $L=2 \mathrm{H}$
We know that, the instantaneous current in the circuit containing $R$ and $L$ in series with battery of emf $E$ is
$$
I=I_0\left(1-e^{\frac{-R}{L} t}\right)
$$
where, maximum current at steady state $I_0=E / R$
$$
\therefore \quad I=\frac{E}{R}\left(1-e^{\frac{-R}{L} t}\right)
$$
Now, the rate of increase of current,
$$
\frac{d I}{d t}=\frac{E}{R}\left[0-\left(\frac{-R}{L}\right) e^{\frac{-R}{L} t}\right]=\frac{E}{L} e^{\frac{-R}{L} t}
$$
At $(t=0)$ initial condition,
$$
\left.\frac{d I}{d t}\right|_{t=0}=\frac{E}{L} e^0=\frac{E}{L}=\frac{6}{2}=3 \mathrm{~A} / \mathrm{s}
$$
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