Question

A linear wire wound potentiometer has a total resistance of $10k\Omega$ and the maximum power dissipation is limited to 50mW. Assuming maximum premissible excitation voltage has been applied.Find out the output voltage for a input displacement of 12.5 mm. The total length of POT is 50 mm. Output voltage in (Volts)

• A. 5.59

Answer 1

The correct option is A 5.59
$\text{Maximum permissible excitation voltage}$

$V_s=\sqrt{P_{max}\times R_P}$

$P_{max}=\text{Power dissipation}=50mW$

$R_P=\text{Potentiometer Resistance}=10k\Omega$

$V_s=\sqrt{50\times {10}^{-3}\times 10\times {10}^3}=\sqrt{500}=22.36V$

$\text{Output Voltage},V_o=\frac{x_i}{x_t}\times V_s$

$x_i=\text{Input Length}=12.5mm$

$x_t=\text{Total Length}=50mm$

$\therefore V_o=\frac{12.5}{50}\times 22.36$

$V_o=5.59Volt$