Part A :The voltmeter reads the voltage which is equal to the voltage drop across the resistance Rv
Vv = I Rv ------------------(1)
here "I" is the current in the circuit
Equivalent resistance of the circuit is
Req = Rb + Rv ( Rb , Rv are in series )
Req = (9.10 + 470 ) = 479.10 .
Now the current in the circuit is I = EMF / (Req)
I = 90.0 V / ( 479.10 )
I = 0.1879 A.
now from (1) substituting all values we get
Vv = ( 0.1879 A ) ( 470 )
Vv = 88.31 V
Part B : Percent error of the reading of the EMF of a battery is
E = (V - Vv ) / V { E= Error , V =EMF = I ( Rb + Rv ) and Vv = I Rv }
E = [ I ( Rb + Rv ) - I Rv ] / [ I ( Rb + Rv )]
E = [ Rb / (Rb + Rv ) ]
Rb E + Rv E = Rb
Rb (1-E) = E Rv
Rb / Rv = E / (1-E)
(Rb / Rv)max = 0.036
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