Question

If a 9 V battery with 4 Ω contact resistance is used and the relay has 80 Ω and the wire has 10 Ω/mile, what is the maximum telegraph distance which will result in a 50 mA current through the relay circuit loop?

Answer 1

Answer:

Telegraph distance will be 9.6 mile

Explanation:

We have given the voltage V = 9 volt

Contact resistance = 4 ohm

Relay resistance = 80 ohm

And wire has a resistance = 10 ohm/mile

We have given the current = 50 mA =0.05 A

According to ohm's law $R=\frac{V}{I}=\frac{9}{0.05}=180ohm$

So the resistance of wire = 180-4-80=96 ohm

So the length of the telegraph distance will be $\frac{96}{10}=9.6mile$