Question

. An ideal parallel plate capacitor of capacitance C has circular plates located at z = 0 and z = d respectively. The medium between the plates is a linear, homogeneous, isotropic dielectric of dielectric constant K. The capacitor is connected to a resistance R in series, and a voltage V is applied to the circuit. The charge q on the capacitor plates increases with time according to q = C V (1-e−t/RC). Find the magnitude of the
magnetic field H inside the dielectric.

Answer 1

As per the given question,

Number of turns in the primary coil $(N_1)=500$

Number of turns in the secondary coil $(N_2)=200$

Resistance in the primary coil $(R_1)=0.3\Omega$

Resistance in the secondary coil$(R_2)=0.02\Omega$

Leakage reactance in the primary coil $(X_1)=2\Omega$

Leakage reactance in the secondary coil$(X_2)=0.05\Omega$

Let R_1' be the resistance of the resistance of primary referred to as secondary ,

$\Rightarrow R_1' =R_1(\frac{N_2}{N_1})^2$

$\Rightarrow R_1'=0.3\times (\frac{200}{500})^2=\frac{0.3\times 4}{25}\Omega=0.048\Omega$

Let R_2' be the resistance of the secondary referred to as primary,

$\Rightarrow R_2'=R_2(\frac{N_1}{N_2})^2=0.02\times (\frac{500}{200})^2\Omega$

$=0.02\times 6.28 \Omega =0.1248\Omega$

Let $X_1'$ be the leakage reactance of the primary referred to as secondary,

$\Rightarrow X_1'=X_1(\frac{N_2}{N_1})^2=2\times (\frac{200}{500})^2=\frac{8}{25}\Omega=0.32\Omega$

Let$X_2'$ be the leakage reactance of the secondary referred to as primary,

$\Rightarrow X_2' =X_2(\frac{N_1}{N_2})^2=0.05\times (\frac{500}{200})^2=0.05\times\frac{25}{4}=0.3125\Omega$

a) Equivalent resistance and reactance referred to as primary,

$R_{eq1}=R_1+R_2'=(0.3+0.1248)\Omega =0.4248\Omega$

$X_{eq1}=X_1+X_2'=2\Omega+0.3125\Omega =2.3125\Omega$

b) Equivalent resistance and reactance referred to as secondary coil,

$R_{eq2}=R_2+R_1'=(0.02+0.048)\Omega =0.068\Omega$

$X_{eq2}=X_2+X_1'=0.05\Omega+0.32\Omega =0.37\Omega$

Equivalent impedance referred to as primary side,

$z=\sqrt{R_{eq1}^2+X_{eq1}^2}=\sqrt{0.4248^2+2.3125^2}=2.35\Omega$