Answer to Question #123866 in Electricity and Magnetism for Perry

Question #123866
. 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.
1
Expert's answer
2020-06-29T14:15:38-0400

As per the given question,

Number of turns in the primary coil (N1)=500(N_1)=500

Number of turns in the secondary coil (N2)=200(N_2)=200

Resistance in the primary coil (R1)=0.3Ω(R_1)=0.3\Omega

Resistance in the secondary coil(R2)=0.02Ω(R_2)=0.02\Omega

Leakage reactance in the primary coil (X1)=2Ω(X_1)=2\Omega

Leakage reactance in the secondary coil(X2)=0.05Ω(X_2)=0.05\Omega

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

R1=R1(N2N1)2\Rightarrow R_1' =R_1(\frac{N_2}{N_1})^2

R1=0.3×(200500)2=0.3×425Ω=0.048Ω\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,

R2=R2(N1N2)2=0.02×(500200)2Ω\Rightarrow R_2'=R_2(\frac{N_1}{N_2})^2=0.02\times (\frac{500}{200})^2\Omega

=0.02×6.28Ω=0.1248Ω=0.02\times 6.28 \Omega =0.1248\Omega

Let X1X_1' be the leakage reactance of the primary referred to as secondary,

X1=X1(N2N1)2=2×(200500)2=825Ω=0.32Ω\Rightarrow X_1'=X_1(\frac{N_2}{N_1})^2=2\times (\frac{200}{500})^2=\frac{8}{25}\Omega=0.32\Omega

LetX2X_2' be the leakage reactance of the secondary referred to as primary,

X2=X2(N1N2)2=0.05×(500200)2=0.05×254=0.3125Ω\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,

Req1=R1+R2=(0.3+0.1248)Ω=0.4248ΩR_{eq1}=R_1+R_2'=(0.3+0.1248)\Omega =0.4248\Omega

Xeq1=X1+X2=2Ω+0.3125Ω=2.3125ΩX_{eq1}=X_1+X_2'=2\Omega+0.3125\Omega =2.3125\Omega

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

Req2=R2+R1=(0.02+0.048)Ω=0.068ΩR_{eq2}=R_2+R_1'=(0.02+0.048)\Omega =0.068\Omega

Xeq2=X2+X1=0.05Ω+0.32Ω=0.37ΩX_{eq2}=X_2+X_1'=0.05\Omega+0.32\Omega =0.37\Omega

Equivalent impedance referred to as primary side,

z=Req12+Xeq12=0.42482+2.31252=2.35Ωz=\sqrt{R_{eq1}^2+X_{eq1}^2}=\sqrt{0.4248^2+2.3125^2}=2.35\Omega


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