Question

The coil of a moving coil voltmeter is 40 mm long and 30 mm wide and has 100 turns on it. The control spring exerts a torque of $240\times {10}^{-6}N-m$ when the deflection is 100 divisions on full scale. If the flux density of the magnetic field in the airgap is $1.0Wb/m^2$, the resistance that must be put in series with the coil to give one volt per division is ________$\Omega$. (Assume resistance of the voltmeter coil is negligible)

• A. $500k\Omega$
• B. $50k\Omega$
• C. $5k\Omega$
• D. $1k\Omega$

Answer 1

The correct option is B $50k\Omega$

Controlling torque at full scale deflection

$T_c=240\times {10}^{-6}N-m$

Deflecting torque at full scale deflection

$T_d=\left(NBld\right)I=100\times 1\times 40\times {10}^{-3}\times 30\times {10}^{-3}I$

$=120\times {10}^{-3}IN-m$

$\text{At final steady position,}{T}_d=T_c$

$120\times {10}^{-3}I=240\times {10}^{-6}$

$\therefore$ Current at full scale deflection,

$I=2\times {10}^{-3}A=2mA$

Let the resistance of the voltmeter circuit be R

$\therefore$ Voltage across the instrument
$=2\times {10}^{-3}R$

This produces a deflection of 100 divisions

$\therefore$ $\text{Volts per division =}\frac{2\times {10}^{-3}R}{100}$

This value should be equal to 1 in order to get 1 volt per division

$\therefore$ $\frac{2\times {10}^{-3}R}{100}=1$

$R=50000\Omega =50k\Omega$