Question 3-9B1

To determine the necessary series resistance, we apply Ohm's Law, which states that resistance ($R$) equals voltage ($V$) divided by current ($I$). In this case, we need to create a voltmeter where the total resistance in the circuit limits the current to the meter's full-scale deflection current at the desired maximum voltage. The formula is: $R = V / I$ Given: * $V = 200$ volts DC * $I = 50$ microamperes DC ($50 \times 10^{-6}$ Amperes) Substitute the values: $R = 200 \text{ V} / (50 \times 10^{-6} \text{ A})$ $R = 4 \times 10^6 \text{ ohms}$ $R = 4 \text{ megohms}$ This 4 megohm resistor, placed in series with the 50 microampere meter, will allow exactly 50 microamperes to flow when 200 volts are applied across the combination, thus giving a full-scale deflection. Options B, C, and D are incorrect as they would result in different current values for the same applied voltage, or require different voltages for full-scale deflection with the 50 microamp meter.