Question

Find the resistance (in ohms) that must be placed in series with a 10.0-Ω galvanometer (100-μA sensitivity) to allow it to be used as a voltmeter with a 300-V full-scale reading.

(a) 2.97 × 10^5 Ω
(b) 3.14 × 10^5 Ω
(c) 3.28 × 10^5 Ω
(d) 3.42 × 10^5 Ω

Answer 1

To use a 10.0-Ω galvanometer with 100-μA sensitivity as a voltmeter for a 300-V full-scale reading, an additional series resistance of approximately 3.00 × 10^6 Ω, or 3.00 Megohms, is required.

Step-by-step explanation:

To find the resistance that must be placed in series with a 10.0-Ω galvanometer having a 100-μA sensitivity to allow it to be used as a voltmeter with a 300-V full-scale reading, we use Ohm's Law (V = IR).

The galvanometer's full-scale deflection is achieved with a current of 100 μA (which is 0.0001 A), and we want this to correspond to a 300-V potential difference.

By rearranging Ohm's Law to solve for the total resistance, Rtot, necessary to produce this voltage, we get Rtot = V / I = 300 V / 0.0001 A = 3,000,000 Ω.

Since the galvanometer already has a resistance of 10.0 Ω, the additional resistance needed in series, R, is R = Rtot - galvanometer resistance = 3,000,000 Ω - 10 Ω = 2,999,990 Ω, which can be approximated to 3.00 × 106 Ω or 3.00 Megohms.