Differential equation application a series rlc circuit has r...

Differential Equation Application. A series RLC circuit has R = 40 ohms, L = 100mH and C = 50 microfarad. The circuit is connected across a 100 V dc source at t = 0 through a switch. Determine the current 0.02 second after the switch is closed. Assume all initial condition to be zero. Ans.: 45.3 mA

Transcript

00:01 So here in this question we are given a series rlc circuit and from this we can say that let's say we are having this circuit this is resistance when resistance is connected to inductor and then that inductor is connected to capacitor both all these three are in series which are then further connected to a power source which is let's say we of dc so from here we can say that the apply we are applying here the kvl to the circuit so we are we are v of dc minus of i r minus of l di divided by the dt minus of 1 divided by c of i dt that will be equals to 0 now we are taking taking place transformation on the laplace transformation on the both sides so we have v of dc divided by the s minus r of i of s minus of l s minus of l s which is in solid state minus 1 divided by the c s of i of s that will be close to zero so this term will be close to r plus l of s plus l divided by the c of s multiplied by the i of s this term so here it is so from here i of s will be equals to v of d c divided by the s so i of s will be equals to r of c s plus l c s square plus i divided by the c of s multiply by the i so this is the value of s and from here we can say that the i of s from here will be equals to b of d c divided by the s that will be equals to lc multiply by the s raised to the power 2 plus r divided by the l of s plus 1 divided by the lc which is divided by the c of s so from here i s will be equals to s raised to the power 2 plus r divided by the l of s plus 1 divided by the l of i so from here we can say that i of s will be equals to v of d c divided by the l which is further divided by s rest to the power 2 plus r divided by the l of s plus 1 divided by the l of c lc so this is the value of i from here now we are given the value of v of d c that is equal to 100 volt r is given that is 40 oom and inductance is given that is equal to 100 million henry so from here c will is capacitance which is 50 so i of s from here putting all these values in the formula will be equals to thousand divided by s plus 200 square of this term plus 400 square so from here this is the value of i of s and from here we can say that the current i of s from here will be equals to 2 .5 multiplied by the 400 divided by the s plus 200 square of this term plus 400 square...

Question

Differential Equation Application. A series RLC circuit has R = 40 ohms, L = 100mH and C = 50 microfarad. The circuit is connected across a 100 V dc source at t = 0 through a switch. Determine the current 0.02 second after the switch is closed. Assume all initial condition to be zero. Ans.: 45.3 mA

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