Jim Guinn’s PHYS2212 Assignment #11
1.
In the figure, E = 6V and R = 0.5
Ω
.
The rate of Joule heating in R is 8W .
a)
What is the current in the circuit?
The Joule heating rate is P = I
2
R , thus I = (P/R)
1/2
=
4A .
b)
What is the potential difference across R?
Similarly, P = V
2
/R or V = (PR)
1/2
=
2V .
c)
What is r?
The voltage drop across R is 2V ; therefore, since the
total voltage drop in the circuit is 6V , the voltage across r is 4V .
Hence, r = V/I = 4V / 4A =
1
Ω
.
2.
A 0.12
μ
F capacitor, in series with a resistor, is given a charge Q
o
.
After 4s its charge is
Q
o
/2 .
What is the resistance of the resistor?
It is given that Q = Q
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 Fall '11
 JimGuinn
 Voltage drop, Joule Heating, total voltage drop, Joule heating rate

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