Question

A series RL circuit with R=30 ohm and L=15H has a constant voltage E =60V applied at t=0 as shown in figure. Determine the current i, the voltage across resistor and inductor.​

Answer 1

Answer:

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Answer 2

Given:

Resistance $R=30 \ \Omega$

Inductance $L=15\ H$

constant voltage $E=60\ V$

The voltage applied at $t=0$

To Find:

The current i, the voltage across the resistor and inductor

Solution:

Use KVL in the circuit

$L\dfrac{dI}{dt}+IR=60\\\dfrac{dI}{dt}+\dfrac{R}{L}I=60\\\dfrac{dI}{dt}+2I=60$

Integrate both sides

$I(t)e^{2t}=\int\ 60\ e^{2}\ dt + C_{o}\\I(t)e^{2t}=30e^{2t}+C_{o}\\\Rightarrow I(t)=30+C_{o}e^{-2t}\\I(t=0)=0\\\Rightarrow C_{o}=-30\\I(t)=30(1-e^{-2t})$

Voltage across Resistance

$V_{R}(t)=I(t)\times R\\V_{R}(t)= 30\times30[1-e^{-2t}]\\V_{R}(t)=900 (1-e^{-2t})$

Voltage across Inductance

$V_{L}(t)=L\dfrac{dI}{dt}\\=15(30)(2e^{-2t})\\V _{L}(t)=900\ e^{-2t}$