Q1Electronic Measurement & Instrumentation

Given parameters: - Resistance of standard resistor $R_s=0.3\Omega$ - Voltage drop across standard resistor $V_R=0.612\angle 12^{\circ }\text{ V}$ - Voltage across test coil via $100/1$ ratio box $V_{box}=0.781\angle 50^{\circ }48^{\prime }\text{ V}$ - Supply Frequency $f=50\text{ Hz}$

Step 1: Calculate the circuit current ($I$) Since the standard resistor is in series with the coil, the same current flows through both. $I=\frac{V_R}{R_s}=\frac{0.612\angle 12^{\circ }}{0.3}=2.04\angle 12^{\circ }\text{ A}$

Step 2: Calculate actual voltage across the test coil ($V_{coil}$) The reading is taken through a $100/1$ ratio box, so the actual voltage is 100 times the reading. Convert $48^{\prime }$ to decimal degrees ($48/60=0.8^{\circ }$). $V_{coil}=100\times 0.781\angle 50.8^{\circ }=78.1\angle 50.8^{\circ }\text{ V}$

Step 3: Calculate the Impedance of the coil ($Z_{coil}$) $Z_{coil}=\frac{V_{coil}}{I}=\frac{78.1\angle 50.8^{\circ }}{2.04\angle 12^{\circ }}$ $Z_{coil}=\frac{78.1}{2.04}\angle (50.8^{\circ }-12^{\circ })=38.284\angle 38.8^{\circ }\Omega$

Step 4: Extract Inductive Reactance ($X_L$) and Inductance ($L$) Convert the polar impedance to rectangular form ($R+j{X}_L$): $X_L=|Z_{coil}|\sin (\theta )=38.284\times \sin (38.8^{\circ })$ $X_L=38.284\times 0.6266=23.988\Omega$

Since $X_L=2\pi fL$: $L=\frac{X_L}{2\pi f}=\frac{23.988}{2\times \pi \times 50}=\frac{23.988}{314.159}=0.07635\text{ H}=76.35\text{ mH}$