Question 3.6: a) A 50 mV, 1 mA d’Arsonval movement is to be used in a volt...

a) Full-scale deflection requires 50 mV across the meter movement, and the movement has a resistance of 50Ω. Therefore we apply Eq. 3.22 v_{2}=i R_{2}=v_{s}\frac{R_{2}}{R_{1}+R_{2}} with R_{1}=R_{v},R_{2}=50,v_{s}=150,and v_{2}=50 mV:

50\times 10^{-3}=\frac{50}{R_{v}+50}\left(150\right) .

Solving for R_{v}gives

R_{v}=149,950\Omega.

b) For a full-scale reading of 5 V,

50\times 10^{-3}=\frac{50}{R_{v}+50}\left(5\right) ,

or

R_{v}=4950\Omega.

c) If we Let R_{m} represent the equivalent resistance of the meter,

R_{m}=\frac{150 V}{10^{-3} A}=150,000\Omega ,

or, alternatively,

R_{m}= 149,950 + 50 = 150,000\Omega ,

d) Then,

R_{m}=\frac{5 V } {10^{-3}A}=5000\Omega ,

or, alternatively,

R_{m}= 4950 + 50 = 5000\Omega .