Question

The switch $S$ in the circuit is connected with point a for a very long time, then it is shifted to position $b$. The resulting current through the inductor is shown by curves in the graph for four sets of values for the resistance $R$ and inductance $L$ (given in column 1). Which set corresponds with which curve?



$\begin{array}{cc}\text{Column I}& \text{Column II}\\ i.R=R_0\text{and}L=L_0& a.I\\ ii.R=2R_0\text{and}L=L_0& b.II\\ iii.R=R_0\text{and}L=2L_0& c.III\\ iv.R=2R_0\text{and}L=2L_0& d.IV\end{array}$

• A. $i-b,ii-c,iii-a,iv-d$
• B. $i-c,ii-b,iii-a,iv-b$
• C. $i-b,ii-a,iii-a,iv-b$
• D. $i-c,ii-a,iii-d,iv-b$

Answer 1

The correct option is D $i-c,ii-a,iii-d,iv-b$

When the switch is connected with $a$ for a long time, current in the circuit would be $\frac{E}{R}$.
When switch is connected to $b,$
$iR+L\frac{di}{dt}=0$
${\int }_{i_0}^i\frac{di}{i}={\int }_0^t\frac{-R}{L}dt$
$i=i_0\left(e^{\frac{-Rt}{L}}\right)$
For $\left(i\right),$
$R=R_0,L=L_0$
At $t=0,i_1=\frac{E}{R_0}\text{and}\tau =\frac{L_0}{R_0}$
$i_1=\frac{E}{R_0}\left(e^{\frac{-R_{0}t}{L_0}}\right)$
For $\left(ii\right),$
$R=2R_0,L=L_0$
At $t=0,i_2=\frac{E}{2R_0}\text{and}\tau =\frac{L_0}{2R_0}$
$i_2=\frac{E}{2R_0}\left(e^{\frac{-2R_{0}t}{L_0}}\right)$
For $\left(iii\right),$
$R=R_0,L=2L_0$
At $t=0,i_3=\frac{E}{R_0}\text{and}\tau =\frac{2L_0}{R_0}$
$i_3=\frac{E}{R_0}\left(e^{\frac{-R_{0}t}{2L_0}}\right)$
For $\left(iv\right),$
$R=2R_0,L=2L_0$
At $t=0,i_4=\frac{E}{2R_0}\text{and}\tau =\frac{L_0}{R_0}$
$i_4=\frac{E}{2R_0}\left(e^{\frac{-R_{0}t}{L_0}}\right)$
Hence at $t=0,i_1=i_3>i_2=i_4$
Also less the time constant faster is the decay,
As ${\tau }_3>{\tau }_1,$ hence $1$ will decay faster than $3$.
As ${\tau }_4>{\tau }_2,$ hence $2$ will decay faster than $4$.
On comparision with given graphs,
$i.\to c.,ii.\to a.,iii,\to d.,iv.\to b$