Answer: the scalar magnitude on lead I is 0.5 mV
Explanation:
Given that;
scalar magnitude on lead II = 1 mV
scalar magnitude on lead III = 0.5 mV
the scalar magnitude on lead I = ?
we know that;
Lead I Voltage = LA - RA -----------let this be equation 1
where LA is left arm electrode and RA is right am electrode
Also
Lead II = LL - RA
where LL is the left leg of electrode
we substitute
1 mV = LL - RA ---------------------let this be equation 2
Again
Lead III = LL - LA
we substitute
0.5 mV = LL - LA ------------------let this be equation 3
now subtract equation 3 and 2
1 mV - 0.5 mv = LL - RA - (LL - LA)
0.5 mV = LL - RA - LL + LA
0.5 mV = -RA + LA
0.5 mV = LA - RA
now taking a look at our equation 1 ( Lead I Voltage = LA - RA )
hence, Lead I Voltage = LA - RA = 0.5 mV
Therefore the scalar magnitude on lead I is 0.5 mV
