Electrical and Electronic Measurements Test 5

Voltmeter reading (V) = 50 V

Milli ammeter reading (I) = 100 mA

\(Measrued\;resistance\;\left( {{R_m}} \right) = \frac{V}{I} = \frac{{50}}{{100 \times {{10}^{ - 3}}}} = 500\;{\rm{\Omega }}\) 

Ammeter is connected in to load side. So, the error will be voltage drop across ammeter.

True voltage drop across resistance = Voltmeter reading – Voltage drop across ammeter

= 50 – (1) (100 × 10^-3) = 49.9 V

\(True\;value\;of\;resistance\;\left( {{R_T}} \right) = \frac{{49.9}}{{100 \times {{10}^{ - 3}}}} = 499\;{\rm{\Omega }}\) 

\(Percentage\;error = \frac{{{R_m} - {R_T}}}{{{R_T}}} \times 100\) 

\( = \frac{{500 - 499}}{{499}} \times 100 = 0.2\% \) 

Insulation resistance of the cable is,

\(R = \frac{{0.4343\;t}}{{C\;log\;\left( {\frac{V}{v}} \right)}}\)

\(\Rightarrow R = \frac{{0.4343 \times 60}}{{500 \times {{10}^{ - 6}}\log \left( {\frac{{400}}{{100}}} \right)}}\)