Question

A battery with EMF 90.0 V has internal resistance R_b = 9.10 Ohm. What is the reading V_v of a voltmeter having total resistance R_v = 470 Ohm when it is placed across the terminals of the battery? Express your answer with three significant figures. What is the maximum value that the ratio R_b/R_v may have if the percent error in the reading of the EMF of a battery is not to exceed 3.50%? Express your answer with three significant figures.

Answer 1

Part A :The voltmeter reads the voltage which is equal to the voltage drop across the resistance R_v

V_v = I R_v   ------------------(1)

here "I" is the current in the circuit

Equivalent resistance of the circuit is

R_eq = R_b + R_v   ( R_b , R_v are in series )

R_eq = (9.10 $\Omega$ + 470 $\Omega$ ) = 479.10 $\Omega$ .

Now the current in the circuit is I = EMF / (R_eq)

I = 90.0 V / ( 479.10 $\Omega$)

I = 0.1879 A.

now from (1) substituting all values we get

V_v = ( 0.1879 A ) ( 470 $\Omega$ )

V_v = 88.31 V

Part B : Percent error of the reading of the EMF of a battery is

E = (V - V_v ) / V { E= Error , V =EMF = I ( R_b + R_v ) and V_v = I R_v }

E = [ I ( R_b + R_v ) -  I R_v ] / [ I ( R_b + R_v )]

E = [ R_b / (R_b + R_v ) ]

R_b E + R_v E = R_b

R_b (1-E) = E R_v

R_b / R_v = E / (1-E)

(R_b / R_v)_max  = 0.036