Question

When the current in the portion of the circuit shown in the figure is 4 A and increasing at the rate of 4 As-1, the measured potential difference VPQ=16V. however when the current is 2A and decreasing at the rated of 1 As-1, the measured potential difference VPQ=5 V. then

• A. $L=1H,R=3\Omega$
• B. $L=2H,R=4\Omega$
• C. $L=3H,R=1\Omega$
• D. $L=4H,R=2\Omega$

Answer 1

The correct option is C $L=3H,R=1\Omega$

Case-1 $I=4A$

$\frac{dI}{dt}=4A/s$

$V=V_{P}Q=16V$

Case-2 $I=2A$

$\frac{dI}{dt}=-1A/s$

$V=-V_{PQ}=5V$

Case-1

$16=L\frac{dI}{dt}+RI$

$=L\times 4+R\times 4$

or, $16=4(L+R)$

or, $L+R=4⟶\left(1\right)$

Case-2

$5=L\frac{dI}{dt}+RI$

or, $5=L\times (+1)+2R$

or, $5=+L+2R$

now, $L+R=4⟶\left(1\right)$ from equation(1)

or, $L=4-1$

$=3H$