Question #224597

a) asi d’ rsonval meter movement with an internal resistan e Rm = 100 Ω and a full s ale

current of Im = 1 mA is to be converted into a multiple-range d.c voltmeter with ranges of

0 – 10 V, 0 – 50V, 0 – 250 V and 0 – 500 V. Find the values of the various resistances using

the potential divider arrangement.


1
Expert's answer
2021-08-11T09:54:18-0400

Given that: Internal resistance of the meter Rm=100 ΩR_{m} = 100\space \Omega


Full scale deflection of current IFSD =Im=1 mA= I_{m} = 1\space mA


Here, we know that the lowest range of the voltmeter is 0–10 V. Therefore, this range must correspond to the resistance nearest to the meter,


The total resistance RTR_{T} for this voltage range can be obtained as:

RT=VIFSD=101×103=10 KΩR_T=\dfrac{V}{I_{FSD}}=\dfrac{10}{1\times10^{-3}}=10\space K\Omega

R4=RTRm=10000Ω100Ω=9900ΩR _4 ​ =R_T ​ −R _m ​ =10000Ω−100Ω=9900Ω


Similarly, for 0–50 V range, the total resistance RTR_{T} becomes:

RT=VIFSD=501×103=50 KΩR_T=\dfrac{V}{I_{FSD}}=\dfrac{50}{1\times10^{-3}}=50\space K\Omega


Similarly, for 0–250 V range, the total resistance RTR_{T} becomes:

RT=VIFSD=2501×103=250 KΩR_T=\dfrac{V}{I_{FSD}}=\dfrac{250}{1\times10^{-3}}=250\space K\Omega


Similarly, for 0–500 V range, the total resistance RTR_{T} becomes:

RT=VIFSD=5001×103=500 KΩR_T=\dfrac{V}{I_{FSD}}=\dfrac{500}{1\times10^{-3}}=500\space K\Omega


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