Question 1.28: A 9 V power supply is to be tested with a 39 Ω load resistor......
A 9 \mathrm{V} power supply is to be tested with a 39 \Omega load resistor. If the resistor has a tolerance of 10 \% find:
(a) the nominal current taken from the supply;
(b) the maximum and minimum values of supply current at either end of the tolerance range for the resistor.
(a) If a resistor of exactly 39 \Omega is used the current will be:
I=V / R=9 \mathrm{V} / 39 \Omega=231 \mathrm{~mA}
(b) The lowest value of resistance would be (39 \Omega-3.9 \Omega)=35.1 \Omega. In which case the current would be:
I=V / R=9 \mathrm{~V} / 35.1 \Omega=256.4 \mathrm{~mA}
At the other extreme, the highest value would be (39 \Omega+3.9 \Omega)=42.9 \Omega. In this case, the current would be:
I=V / R=9 \mathrm{~V} / 42.9 \Omega=209.8 \mathrm{~mA}
The maximum and minimum values of supply current will thus be 256.4 \mathrm{~mA} and 209.8 \mathrm{~mA}, respectively.