3 en el circuito de la siguiente figura la corriente de la...

3. En el circuito de la siguiente figura la corriente de la fuente está dada por la ecuación $i_F(t)$, si la inductancia es $L_1$ determine: a. La ecuación del voltaje $V_o$ en el inductor: $V_o(t)=$ b. La ecuación de la potencia en función del tiempo en el inductor $P_L(t)=$ c. La ecuación de la energía almacenada en el inductor en función del tiempo $W_L(t)=$ $i_F(t) = kte^{at}$ A L= 16mH k= 8 a= -13

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00:01 Hello everyone so according to this given problem for this given rl circuit means the question is related to rl circuit right which is given the potential of e right and we need to find the current i at t is greater than zero so let's say this is our rl circuit this is the resistor and this is our inductor l right and it is connected to potential e right so the current at t greater than 0 is given as i t is equals to e by r right one minus e power minus r t by l so this is the equation for the current at time t greater than 0 right so let's put these values so for normal case we have e as variable and the value of r is 8 and 1 minus e power minus 8 t by l and value of l is basically 0 .5 so when we put l is equal to 0 .5 so we will end up getting e power minus 16 t so this is the equation for current time t greater than 0 now for the maximum value of current what we will do we will take time t tends to infinity right means for the larger time for like for the larger time the i t means the saturated value of current will goes like e by 8 1 minus e power minus 16 so when we put the value of t is equal to infinity so we will get e power minus infinity which will give us e by 8 because e power minus infinity is 0 okay so this will give us i t means the saturated value of current as e by 8 only right so let's let let's put the value of e so first case when e is equal to 64 so the value of i t will be 64 by 8 that implies the current maximum saturated current for is equal 64 is 8 ampere right and uh for uh second case where e is equals to 8 t e power minus 16 t the value of i t will be uh this is a 8 t e power minus 16 t divided by 8 so 8 and 8 will cancel out each other and we will get t e power minus 16 t so the value of in this case is equal to 8 t e power minus 16 t the i t comes out to be t multiplied e power minus 16 t so as you put the value t tends to infinity right eventually you will get the i t tends to be 0 because when you put uh t tends to infinity then e power minus infinity will give you zero hence your final answer will be zero now similarly for this other case third one where e is equals to 32 e power minus 8t so the value of i t comes out to be 32 divided by 8 e power minus 8 t so for the long time we take the limit limit t tends to infinity uh for e power minus 8 t so eventually in this case also we we will get the answer 0...

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