A pure resistance and a pure inductance are connected in series across a 100 volt A.C line. A voltmeter gives same reading whether connected across resistance or inductance. It does read ........ V.

Options

A. 50 V
B. 70.7 V
C. 88.2 V
D. 100 V

Answer

B

Explanation

If both the component connected in parallel to the voltmeter, both shows same voltage.
`V_"rms"`=100 V
`therefore V_m =sqrt2xxV_"rms"`
=1.414 x 100
=141.4
Applying Kirchoff.s second law to the circuit
`V_m=V_1+V_2`
`therefore V_m=2V_1` or `2V_2 " " [because V_1=V_2]`
`therefore 141.4 =2V_1`
`therefore V_1=70.7 V` or `V_2`=70.7 V
Second Method :
Here `V_(rms(R))=V_(rms(L))`
`therefore V_(rms)^2=V_R^2+V_L^2`
`therefore V_(rms)^2 =V_R^2+V_R^2 " " [because V_R=V_L]`
`therefore (100)^2 =2V_R^2`
`therefore V_4=100/sqrt2 =70.7 V = V_L`