Question

When the current in a certain inductor coil is 5.0 A and is increasing at the rate of 10.0 A/s, the potential difference across the coil is 140 V . When the current is 5.0 A and deacreasing at the rate of 10.0 A/s, the potential difference is 60 V . Find resistance of coil.

Answer 1

Given that,

Current $i=5A$

Potential difference$V=140V$

Now, case (I)

$V_{total}=V_I+V_r$

$V_{total}=L\frac{di}{dt}+iR$

$140V=L\left(10\right)+5\times R$

Now, case (II)

$V_{total}=L\left(\frac{di}{dt}\right)+iR$

$60V=L(-10)+5\times R$

Now, subtract case (II) from case (I)

$140V-60V=L\left(10A/s\right)-L(-10A/s)$

$80V=L\left(20A/s\right)$

$L=4H$

Now, substitute the value of L in equation (II)

$60V=4H(-10A/s)+5\times R$

$5\times R=60V+40HA/s$

$5\times R=60+40$

$R=\frac{100}{5}$

$R=20\Omega$

Hence, the resistance is $20\Omega$