An inductor -coil of inductance 17 mH is constructed from a cpper wire of length 100 m and cross sectional area 1 mm. caluclate the time constant of the circuit if this inductor is joined across an ideal battery The resistivity of copper $= 1.7 \times 10^{- 8} Ω m .$

Open in App

Solution

$L = 17 m H, l = 100 m, A = 1 m m^{2} = 1 \times 10^{- 6} m^{2} ∴ f c v = 1.7 \times 10^{- 8} Ω - m R = \frac{f c v L}{A} = \frac{1.7 \times 10^{- 2} \times 100}{1 \times 10^{- 6}} = 1.7 Ω i = \frac{L}{R} = \frac{17 \times 10^{- 3}}{1} .7 = 10^{- 2} S e c = 10 s e c .$

Suggest Corrections

Similar questions

Q. An inductor-coil of inductance

20 mH

having resistance of

10 Ω

is joined to an ideal battery of emf

5 V .

Find the emf induced across the inductor at

t = 0 s.

An inductor-coil carries a steady-state current of 2.0 A when connected across an ideal battery of emf 4.0 V. If its inductance is 1.0 H, find the time constant of the circuit.

Q. A copper wire of cross-sectional area

2.0 m m^{2}

, resistivity

= 1.7 \times 10^{- 8} Ω m

, carries a current of

1 A

. The electric field in the copper wire is

Q. What length of a copper wire of cross-sectional area 0.01 mm² will be needed to prepare a resistance of 1 kΩ? Resistivity of copper = 1.7 × 10^–8 Ωm

Q. An inductor coil stores

32 J

of magnetic field energy and dissipates energy as heat at the rate of

320 W

when a current of

4 A

is passed through it. Find the time constant of the circuit when this coil is joined across an ideal battery.

Join BYJU'S Learning Program

An inductor -coil of inductance 17 mH is constructed from a cpper wire of length 100 m and cross sectional area 1 mm. caluclate the time constant of the circuit if this inductor is joined across an ideal battery The resistivity of copper =1.7×10−8Ωm.

L=17mH,l=100m,A=1mm2=1×10−6m2∴fcv=1.7×10−8Ω−mR=fcvLA=1.7×10−2×1001×10−6=1.7Ωi=LR=17×10−31.7=10−2Sec=10sec.

An inductor -coil of inductance 17 mH is constructed from a cpper wire of length 100 m and cross sectional area 1 mm. caluclate the time constant of the circuit if this inductor is joined across an ideal battery The resistivity of copper $= 1.7 \times 10^{- 8} Ω m .$

$L = 17 m H, l = 100 m, A = 1 m m^{2} = 1 \times 10^{- 6} m^{2} ∴ f c v = 1.7 \times 10^{- 8} Ω - m R = \frac{f c v L}{A} = \frac{1.7 \times 10^{- 2} \times 100}{1 \times 10^{- 6}} = 1.7 Ω i = \frac{L}{R} = \frac{17 \times 10^{- 3}}{1} .7 = 10^{- 2} S e c = 10 s e c .$