Question

Design a current divider/shunt resistance that will allow a lmA (full scale) 140 2 (internal resistance) d'Arsonval meter to measure 10mA at full scale (i.e. iunknown 10 mA, imeter 1 mA, Rmeter 140 , Rshunt?). NOTE: You may need to use up to two standard resistor values in series and/or parallel combinations to reach the needed shunt resistance. Lab values are: 10 22, 33, 47, 56, 68, 82, 100, 220, 330, 470, 560, 680, 820, lk, 2.2k, ..., IM2 Calculate the power requirements of all resisters in the design. Include a detailed schematic of your design in your lab book. Using the Rshunt found in Step 1 for the 140 2 d'Arsonval meter, derive equation for iunknown in terms of imeter. This will be used to convert the actual meter readings performed in the lab (imeter) to the current following thru the "new" 10mA meter (iunknown).

Answer 1

Given that I_shunt = I_total - I_meter = 10 m Amp - 1 m Amp = 9 m Amp In parallel voltage must be same I_s times R_sh = I_m R_m R_sh = I_m times R_m/I_sh = 1 m times 140/9 m = 15.556 ohm this was possibly by connecting two 33 ohm in parallel 33 parallel 33 = 15.5 Power dissipated in meter @ 10 m Amp = I_m^2 R_m + I_sh^2 middot R_sh = (1m)^2 (140) + (9m)^2 (15.58) = 1.3955 m watts From the current I_m = I_total times R_sh/R_sh + R_m I_total = [R_sh + R_m/R_sh] I_m I_total = [1 + R_m/R_sh] times I_m

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