= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)
= -0.00225 m
New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
D
18. A 2.0-ohm resistor is connected in a series with a 20.0 -V battery and a three-branch parallel network with branches whose r
esistance are 8.0 ohms each. Ignoring the battery’s internal resistance, what is the current in the batter? Show your work.
2 answers:
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= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)
= -0.00225 m
New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
D
A) The power delivered to the lines is
And the voltage at which the lines work is
Since the power delivered is the product between the voltage and the current:
We can find the current flowing in the lines:
b) The voltage change along each line can be found by using Ohm's law:
c) The power wasted as heat along each line is given by:
And since we have 2 lines, the total power wasted as heat in both lines is
And the voltage at which the lines work is
Since the power delivered is the product between the voltage and the current:
We can find the current flowing in the lines:
b) The voltage change along each line can be found by using Ohm's law:
c) The power wasted as heat along each line is given by:
And since we have 2 lines, the total power wasted as heat in both lines is