Given the following bridge: The power supply is set to 3.50 volts When the left

Question

Given the following bridge: The power supply is set to 3.50 volts When the left probe end is touched to the 14.6 cm position and the night probe is touching the 46.9 cm position on the bridge: a) What is the electric field strength in the wire, in volts/cm? b) What is the potential difference between the two probes? (c) If the power supply was adjusted to a voltage of 5.00 volts, how would you have to move the left probe from the 14.6 cm position to have the meter read 0.60 volts (that is. leaving the right probe fixed)? (to the left, towards the 0 cm mark, or to the night towards the 100 cm mark) - show why this is so. What circuit rule or law explains why an ac adapter plugged into a wall socket gets warm? Be specific.

Explanation / Answer

a)

eletric field = V/d

eletric field = 3.50/1

eletric field = 3.5 V/m

b)

potential difference = 3.5 * (0.469 - 0.146)

potential difference = 1.31 V

c)

for the reading to be 0.6 V

it should be moved towards the 100 cm mark

and the reading will be

0.6 = 5 * (0.469 - x)

x = 0.349 m = 34.9 cm

it should be moved to 34.9 cm

4)

It is due to resistance in the plug

and power = I^2 * R

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