A coil with an inductance of 2.0 H and a resistance of 10 Ωis suddenly connected to an ideal bat

step 1 Hi, in the given problem there is a series LR circuit. So in this series LR circuit, the value o

A coil with an inductance of 2.0 H and a resistance of 10...

Key Concepts

Battery Energy Delivery

This concept involves the work done or power delivered by a voltage source when supplying energy to a circuit. The instantaneous power from a battery is calculated using P = V I, where voltage and current might be time-dependent in a transient state. Understanding this helps in accounting for the energy flow balance in the circuit, as the battery’s energy delivery is distributed between energy stored in the inductor and energy lost as heat in the resistor.

Power Dissipation in a Resistor

In any circuit, resistors convert electrical energy into thermal energy (heat) due to the Joule heating effect. The rate at which energy is dissipated in a resistor is given by P = I²R. This concept is fundamental to understanding energy losses in circuits and is especially important when comparing energy storage versus energy dissipation in transient circuits.

RL Circuit Transients

This concept involves analyzing how currents in a circuit containing both resistance (R) and inductance (L) evolve over time after a sudden change, such as closing a switch. The time-dependent behavior is often characterized by an exponential response with a time constant defined by L/R, which governs how fast the current reaches its steady?state value.

Energy Storage in an Inductor

An inductor stores energy in its magnetic field as current flows through it. The energy stored is given by the expression U = (1/2)L I², and the rate of energy storage can be found by differentiating this expression with respect to time. This process is crucial when analyzing transient responses in inductive circuits since it represents energy being temporarily accumulated rather than dissipated.

Transcript

00:01 Hi, in the given problem there is a series lr circuit.

00:07 So in this series lr circuit, the value of self -inductance of the coil is given as 2 .0 henry.

00:23 The resistance here is 10 -oom.

00:28 The emf applied is e -0 is equal to 100 volt.

00:33 So, the maximum current in the circuit will be given by e0 by r, maximum equilibrium current.

00:44 So here this is 100 volt divided by 10 om or we can say this is 10 ampere and time constant of this circuit will be l by r.

00:59 Means this is for l to henry for r this is 10 oms so finally time constant here comes out to be 0 .2 seconds now we have to find at an instant when the time is 0 .10 second first of all in the first part of the problem we have to find the rate at which the energy is being converted into magnetic field energy of the inductor coil.

01:30 So to find it using the expression for magnetic field energy, ub is equal to half l i square.

01:42 So to find the rate of conversion of this energy we differentiate both sides.

01:47 So this d u b differentiation of u b with respect to time will be half l as a constant and then differentiation of i.

01:59 It will be 2i and then di by d t so canceling this two here it becomes the rate of conversion of magnetic field energy is equal to l i into d i by d t so first of all we should find this di by d t so to find it we use the equation of growth of current in l r circuit and that equation is i is equal to i0 bracket 1 minus e ratio power minus t by tau the time constant so here this d i by d t the differentiation of current will come out to be if we look into this bracket keeping this i not as a constant out here it will be 0 and then differentiation of e rase bar minus t by tau will be e -raged to the power minus t by tau as a whole and then multiplied by the differentiation of this power minus t by tau which will be minus 1 by tau only.

03:12 So here it will come out to be minus i not by tau minus i not by tau into e -raise to the power minus t by tau so putting this value of di by d t and here it will become plus not minus because of these two negative signs so finally this d u b by d t will become equal to l i first of all then d i by d t for di by d t this is i not by tau into e dash bar minus t by tau or we can say this is l i not square now now we will use the value of this i also equation instantaneous current equation for instantaneous current which is keeping this l by and for tau we may use l by r and for i this is i0 bracket 1 minus e to the power minus t by tau bracket closed and then again one more i not so it will become i not square and e r to the power minus t by tau finally so ultimately this equation here comes out to be canceling this l and here this r will go up and and finally for e0 we may use this is r for i not this is e not by r square 1 minus e rates bar minus t by tau and e rate bar minus t by tau so canceling this r from this r square finally we get d u b by d t as e square by r bracket 1 minus e -r -r -r -r -r -r -t bar minus t by tau into e -res -to -bar minus t by tau.

05:27 So finally plugging in all the known values for epsilon means maximum voltage, that is 100 -volt.

05:35 So this is a square of 100.

05:37 Resistance is 10 -oom 1 -e -reshe -the -bar 0 .1, the instant at which we have to find this rate of change of energy.

05:47 And tau we have already found it to be 0 .2 minus out of the bracket that is also the same thing e ratio to the power minus 0 .1 divided by 0 .2 so finally solving this we get it to be 2 .4 into 10 dash par 2 watt this is the rate of change of energy means the rate at which the energy is being converted into the magnetic field energy of the inductor coil...

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