Answer to Question #224597 in Electric Circuits for Unknown346307

Question

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.

Answer 1

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

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

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 $R_{T}$ for this voltage range can be obtained as:

$R_T=\dfrac{V}{I_{FSD}}=\dfrac{10}{1\times10^{-3}}=10\space K\Omega$

$R _4 ​ =R_T ​ −R _m ​ =10000Ω−100Ω=9900Ω$

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

$R_T=\dfrac{V}{I_{FSD}}=\dfrac{50}{1\times10^{-3}}=50\space K\Omega$

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

$R_T=\dfrac{V}{I_{FSD}}=\dfrac{250}{1\times10^{-3}}=250\space K\Omega$

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

$R_T=\dfrac{V}{I_{FSD}}=\dfrac{500}{1\times10^{-3}}=500\space K\Omega$