A single turn (10 mathrm{~K}) potentiometer with rotation angle of (320^{circ}), is shown in Fig. P9.1. The
Question:
A single turn \(10 \mathrm{~K}\) potentiometer with rotation angle of \(320^{\circ}\), is shown in Fig. P9.1. The fixed terminals \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are connected to \(+10 \mathrm{~V}\) and \(-10 \mathrm{~V}\) DC supply respectively. The centre tap \(c\) is grounded. \(\mathrm{V} / \mathrm{rad}\).
(a) Find gain constant (sensitivity) of potentiometer in
(b) What is open circuit output voltage \(\mathrm{V}_{0}\) if wiper is rotated through an angle of \(30^{\circ}\) towards fixed terminal \(b\) from centre tap \(\boldsymbol{c}\) ?
(c) What is open circuit output voltage if wiper is rotated through an angle \(60^{\circ}\) towards fixed terminal \(\boldsymbol{a}\) from centre tap \(\boldsymbol{c}\) ?
(d) If this potentiometer has 10 turns instead of single turn, what is new gain constant (volts/rad) and what is output voltage (open circuit) when wiper is rotated \(64^{\circ}\) from centre tap \(\boldsymbol{c}\) towards \(+10 \mathrm{~V}\).
Step by Step Answer:
a To find the gain constant sensitivity of the potentiometer we need to calculate the voltage change per unit of angular change in radians Since the t...View the full answer
