7. (a) Let
i
be the current in the circuit and take it to be positive if it is to the left in
R
1
.
We use Kirchhoff’s loop rule:
ε
1
–
iR
2
–
iR
1
–
2
= 0. We solve for
i
:
i
RR
=
−
+
=
−
+
=
12
12
6 0
80
050
VV
4.0
A
.
.
..
ΩΩ
A positive value is obtained, so the current is counterclockwise around the circuit.
If
i
is the current in a resistor
R
, then the power dissipated by that resistor is given by
2
PiR
=
.
(b) For
R
1
,
P
1
=
2
1
iR
=
(0.50 A)
2
(4.0
Ω
) = 1.0 W,
(c) and for
R
2
,
P
2
=
2
2
=
(0.50 A)
2
(8.0
Ω
) = 2.0 W.
If
i
is the current in a battery with emf
, then the battery supplies energy at the rate
P =i
provided the current and emf are in the same direction. The battery absorbs energy at the
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Current

Click to edit the document details