A coil of self inductance 10 mH and resistance 0.1 Ω is connected through a switch to a battery of internal resistance 0.9 Ω . After the switch is closed, the time taken for the current to attain 80 % of the saturation value is: (Take ln5 = 1.6)

Question

A coil of self inductance 10 mH and resistance 0.1 Ω is connected through a switch to a battery of internal resistance 0.9 Ω . After the switch is closed, the time taken for the current to attain 80 % of the saturation value is: (Take ln5 = 1.6) (A) 0.103 s (B) 0.016 s (C) 0.002 s (D) 0.324 s

Tardigrade Admin · Accepted Answer

i = i0 (1- e-t/τ) (80/100) i0 =i0 (1-e-t/τ) 0.8 = 1 - e-t/τ e-t/τ = 0.2 = (1/5) - (t/τ) = ln((1/5)) - (t/τ) = -In (5) t= τ. ln(5) = (L/Req) .ln (5) = (10×10-3/(0.1+0.9)) ×1.6 t = 1.6 ×10-2 t = 0.016 s

Q. A coil of self inductance 10mH and resistance 0.1Ω is connected through a switch to a battery of internal resistance 0.9Ω. After the switch is closed, the time taken for the current to attain 80% of the saturation value is : (Take ln5=1.6)

0.103 s

0.016 s

0.002 s

0.324 s

i=i0​(1−e−t/τ) 10080​i0​=i0​(1−e−t/τ) 0.8=1−e−t/τ e−t/τ=0.2=51​ −τt​=ln(51​) −τt​=−In(5) t=τ.ln(5) =Req​L​.ln(5)=(0.1+0.9)10×10−3​×1.6 t=1.6×10−2 t=0.016s

Q. A coil of self inductance $10 m H$ and resistance $0.1 Ω$ is connected through a switch to a battery of internal resistance $0.9 Ω$ . After the switch is closed, the time taken for the current to attain $80%$ of the saturation value is : (Take $l n 5 = 1.6$ )