Question

A coil with an inductance of 2.8 H and a resistance of 12 Ω is suddenly connected to an ideal battery with ε = 89 V. At 0.086 s after the connection is made, what is the rate at which (a) energy is being stored in the magnetic field, (b) thermal energy is appearing in the resistance, and (c) energy is being delivered by the battery?

Answer 1

Answer:

The stored energy is 140.7 watt.

The thermal energy is 62.7 watt.

The delivered energy is 203.4 watt.

Explanation:

Given that,

Inductance = 2.8 H

Resistance = 12 Ω

Potential $\epsilon_{0}=89\ V$

Time = 0.086 s

(a). We need to calculate the energy stored in the magnetic field

Using formula of current

$i=i_{max}(1-e^(\frac{-t}{\tau}))$

Using formula of energy

$U=\dfrac{1}{2}Li^2$

On differentiating

$\dfrac{dU}{dt}=Li\frac{di}{dt}$

$\dfrac{dU}{dt}=L\dfrac{d}{dt}(i_{max}(1-e^(\frac{-t}{\tau}))$

Again differentiating

$\dfrac{dU}{dt}=\dfrac{\epsilon^2}{R}(1-e^{\frac{-t}{\tau}})e^{\frac{-t}{\tau}}$

$\dfrac{dU}{dt}=\dfrac{\epsilon^2}{R}(1-e^{\frac{-\t\times R}{L}})e^{\frac{-t\times R}{L}}$

Put the value into the formula

$\dfrac{dU}{dt}=\dfrac{(89)^2}{12}(1-e^{\dfrac{-0.086\times12}{2.8}})e^{\dfrac{-0.086\times12}{2.8}}$

$\dfrac{dU}{dt}=140.7\ watt$

(b). We need to calculate the thermal energy

Using formula of thermal energy

$P=i^2R$

$P=\dfrac{\epsilon^2}{R}(1-e^{\frac{-t}{\tau}})^2$

Put the value into the formula

$P=\dfrac{89^2}{12}(1-e^{\dfrac{-0.086\times12}{2.8}})^2$

$P=62.7\ Watt$

(c). We need to calculate the delivered energy by the battery

Using formula of energy

$P'=P+\dfrac{dU}{dt}$

$P'=62.7+140.7$

$P'=203.4\ watt$

Hence, The stored energy is 140.7 watt.

The thermal energy is 62.7 watt.

The delivered energy is 203.4 watt.