TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Given, resistance, R=50Ω
supply voltage, V=5V
current, I=50mA=50×10−3A
Time taken, t=0.1s
As we know that,
I=I0(1−e−‌

t

τ

) . . (i)
where,
‌I0=‌ peak current, ‌
‌I=‌ current in coil ‌
‌ and ‌τ=‌ time constant ‌=‌

L

R

‌. ‌
Since, by using Ohm's law,
V‌‌=I0R
⇒‌‌I0‌‌=‌

V

R

=‌

5

50

Put in Eq. (i), we get
⇒50×10−3=‌

5

50

(1−e−‌

0.1

τ

)
⇒‌‌5×10−3+2=1−e−0.1∕τ
⇒‌‌e−0.1∕τ=1−0.5=0.5
Taking log both sides, we get
⇒−‌

0.1

τ

logee‌‌=ln‌

1

2

⇒−‌

0.1

τ

=ln‌l−ln‌2
⇒‌‌‌

0.1

τ

‌‌=ln‌2⇒τ=‌

0.1

ln‌2

=‌

L

R

⇒‌‌L‌‌=‌

0.1R

ln‌2

=‌

0.1×50

ln‌2

‌‌=‌

5

ln‌2