3. Given the following bridge 100cm The power supply is set to3.50 vots. When te

Question

3. Given the following bridge 100cm The power supply is set to3.50 vots. When te left prote edistuted tothe 145 poston and the right probe is touching the 46.9 cm position on the bridge a) What is the electric field strength in the wire, in volts/am? 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 mark, orto te right towards the 100 mark-show why this is so. 4. What circuit rule or law explains why an ac adapter plugged into a wall socket gets warm? Be specific.

Explanation / Answer

3 )

a )

the magnitude of the Electric field is " E "

E = V / d

E = 3.5 / 100

E = 0.035 V/cm

b )

potential difference is V = - E ( 14.6 - 46.9 )

= 1.1305 volts

c )

E = 5 / 100

= 0.05 V/ cm

potential difference = 0.6 = - E ( x - 46.9 )

x = 34.9 cm

4 )

Why AC adapter is getting warm is explained by Ohm's Law

the heat power dissipation is P = i X V

P = i X iR

P = i2 R

so the more resistance is more dissipation in the form of heat...

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