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Assuming R₁=5Ω ,

Initially, \$V_C\ + V_{R_1} = V_{R_2}\$. Since no current flows through C for \$t(0^-)\$ , \$i_{R_2}=1A\$. So, \$V_{R_2}=V_C=2V\$.

For \$t>0\$,

$$E=L\frac{di}{dt} + i\cdot R_1 + \frac{1}{C}\int i dt + V_C(0)$$

Taking Laplace Transform,

$$\frac{E}{s} = L(sI(s)-I_L(0)) + I(s) \cdot R_1 + \frac{1}{Cs}I(s) + \frac{V_C(0)}{s}$$

Here \$I_L(0) = 1A, V_C(0) = 2V\$

Rearranging,

$$\frac{E+LsI_L(0)-V_C(0)}{Ls^2 + s R_1 + \frac{1}{C} } = I(s)$$

Substitute the values and resolve into partial fractions and then take Inverse Transform to get the value of the current.

Edit:

The Laplace Inverse of 1/7 is 1/7δ(t).

Assuming R₁=5Ω ,

Initially, \$V_C\ + V_{R_1} = V_{R_2}\$. Since no current flows through C for \$t(0^-)\$ , \$i_{R_2}=1A\$. So, \$V_{R_2}=V_C=2V\$.

For \$t>0\$,

$$E=L\frac{di}{dt} + i\cdot R_1 + \frac{1}{C}\int i dt + V_C(0)$$

Taking Laplace Transform,

$$\frac{E}{s} = L(sI(s)-I_L(0)) + I(s) \cdot R_1 + \frac{1}{Cs}I(s) + \frac{V_C(0)}{s}$$

Here \$I_L(0) = 1A, V_C(0) = 2V\$

Rearranging,

$$\frac{E+LsI_L(0)-V_C(0)}{Ls^2 + s R_1 + \frac{1}{C} } = I(s)$$

Substitute the values and resolve into partial fractions and then take Inverse Transform to get the value of the current.

Edit:

The Laplace Inverse of 1/7 is 1/7δ(t), which is a function that takes the value 0 for all \$t\neq 0\$.

Assuming R₁=5Ω ,

Initially, V_c + V_R1 = V_R2. Since no current flows through C for t(0^-) , i_R2=1A. So, V_R2=V_C = 2V.

For t>0,

E=L(di/dt) + iR1 + (1/C)∫idt + Vc(0) ;

Taking Laplace Transform,

(E/s)=L(sI(s)-I(0)) + I(s)R1 + (1/Cs)I(s) + Vc(0)/s ;

Here I(0) = 1A, Vc(0) = 2V;

Rearranging,

{ E+LsI(0)-Vc(0) } / {Ls2 + sR1 + (1/C) } = I(s)

Substitute the values and resolve into partial fractions and then take Inverse Transform to get the value of the current.

Edit:

The Laplace Inverse of 1/7 is 1/7δ(t).

Assuming R₁=5Ω ,

Initially, \$V_C\ + V_{R_1} = V_{R_2}\$. Since no current flows through C for \$t(0^-)\$ , \$i_{R_2}=1A\$. So, \$V_{R_2}=V_C=2V\$.

For \$t>0\$,

$$E=L\frac{di}{dt} + i\cdot R_1 + \frac{1}{C}\int i dt + V_C(0)$$

Taking Laplace Transform,

$$\frac{E}{s} = L(sI(s)-I_L(0)) + I(s) \cdot R_1 + \frac{1}{Cs}I(s) + \frac{V_C(0)}{s}$$

Here \$I_L(0) = 1A, V_C(0) = 2V\$

Rearranging,

$$\frac{E+LsI_L(0)-V_C(0)}{Ls^2 + s R_1 + \frac{1}{C} } = I(s)$$

Substitute the values and resolve into partial fractions and then take Inverse Transform to get the value of the current.

Edit:

The Laplace Inverse of 1/7 is 1/7δ(t).

Assuming R₁=5Ω ,

Initially, V_c + V_R1 = V_R2. Since no current flows through C for t(0^-) , i_R2=1A. So, V_R2=V_C = 2V.

For t>0,

E=L(di/dt) + iR1 + (1/C)∫idt + Vc(0) ;

Taking Laplace Transform,

(E/s)=L(sI(s)-I(0)) + I(s)R1 + (1/Cs)I(s) + Vc(0)/s ;

Here I(0) = 1A, Vc(0) = 2V;

Rearranging,

{ E+LsI(0)-Vc(0) } / {Ls2 + sR1 + (1/C) } = I(s)

Substitute the values and resolve into partial fractions and then take Inverse Transform to get the value of the current.

Edit: Reply to your comment:

The Laplace Inverse of 1/7 is 1/7δ(t). So, that's where you're wrong.

Assuming R₁=5Ω ,

Initially, V_c + V_R1 = V_R2. Since no current flows through C for t(0^-) , i_R2=1A. So, V_R2=V_C = 2V.

For t>0,

E=L(di/dt) + iR1 + (1/C)∫idt + Vc(0) ;

Taking Laplace Transform,

(E/s)=L(sI(s)-I(0)) + I(s)R1 + (1/Cs)I(s) + Vc(0)/s ;

Here I(0) = 1A, Vc(0) = 2V;

Rearranging,

{ E+LsI(0)-Vc(0) } / {Ls2 + sR1 + (1/C) } = I(s)

Substitute the values and resolve into partial fractions and then take Inverse Transform to get the value of the current.

Edit:

The Laplace Inverse of 1/7 is 1/7δ(t).