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(B) The circuit consists of a battery of $5\,V$,an inductor of $10\,mH$,and a variable resistor $R_H$. According to Kirchhoff's voltage law,the equati

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$i > 0.5\,A$

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(B) The circuit consists of a battery of $5\,V$,an inductor of $10\,mH$,and a variable resistor $R_H$. According to Kirchhoff's voltage law,the equation for the circuit is $V = iR + L\frac{di}{dt}$. If the resistance $R$ were constant at $10\,\Omega$,the steady-state current would be $i = \frac{V}{R} = \frac{5}{10} = 0.5\,A$. However,the resistance is increasing,which means $\frac{dR}{dt} > 0$. As the resistance increases,the current $i$ tends to decrease. According to Lenz's law,the inductor opposes this change by inducing an electromotive force $(EMF)$ that tries to maintain the current. The induced $EMF$ is given by $\varepsilon = -L\frac{di}{dt}$. Since the current is decreasing,$\frac{di}{dt}$ is negative,making the induced $EMF$ positive in the direction of the main current. This induced $EMF$ acts as a source that adds to the current,causing the actual current to be higher than the value it would have if the resistance were constant. Therefore,$i > 0.5\,A$.

The resistance in the following circuit is increased at a particular instant. At this instant,the value of resistance is $10\,\Omega$. The current in the circuit will be

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