Physics 1302W.500, Week 9
Scott Fallows
(Dated: March 24, 2008)
I.
HEAT GENERATION IN A DC CIRCUIT
A second 12 V battery is added to a sensitive low-temperature device. Determine the
heat generated by the 20 Ω resistor in the modified circuit below.
Solution. Label the resistors: R1 = 100 Ω, R2 = 20 Ω, R3 = 40 Ω, R4 = 200 Ω. To find
the heat generated by R2 , we need to find the current through it and then use it to find the
rate of energy being dissipated in the resistor, P = I 2 R. This is easier if we first combine
the parallel resistors into R = R2 R3 /(R2 + R3 ), find the current through R, and then use
the current divider relation to find the fraction of this current going through R2 .
The equivalent circuit now has two nodes (one independent equation from the Kirchhoff
Junction Rule) and two independent loops (two independent equations from the Kirchhoff
Loop Rule). Label the current going clockwise in the left branch I1 , the current going down
in the middle branch I, and the current going clockwise in the right branch I3 . Then the
junction rule gives us
I1 = I + I3
(1)
Now use the loop rule on the left (clockwise) and right (counter-clockwise) sub-loops to find
V − IR − I1 R1 = 0
(2)
V + I3 R4 − IR = 0
(3)
Substitute (1) into (2), then solve (3) for I3 and substitute it back into the modified (2).
This gives
IR − V
V − IR − I +
R4
R1 = 0
which we multiply by R4 and solve for I to get
I=
V (R1 + R4 )
= 150 mA
RR1 + RR4 + R1 R4
Now we need to find the amount of current going through R2 . Expand the middle branch
back to two resistors in parallel. Then the current divider relation gives
I2 =
R3
I = 100 mA
R2 + R3
The dissipated power is P = I22 R2 = 0.2 W.
II.
POWER RATING FOR RESISTORS IN A DC CIRCUIT
The resistors in the following circuit are rated at 0.5 W, meaning they burn up if more
than 0.5 W of power passes through them. Will the 100 Ω resistor burn up?
Solution First redraw the circuit so that the upper-left triangular loop is a rectangular
loop on the left and the lower-right triangular loop is a rectangular loop on the right;
this make the problem easier to visualize. Now the problem is essentially the same as the
previous, with slightly different circuit elements present. Again, we need to find the current
through a resistor to determine the power it dissipates.
Label the components: R1 = 100 Ω, R2 = 133 Ω, R3 = 200 Ω, V1 = 9 V, V2 = 6 V. Call
I1 the clockwise current in the left branch, I2 the downward current in the middle branch,
and I3 the clockwise current in the right branch. Consider the node to the right of R1 . The
Kirchhoff Junction Rule gives
I2 = I1 − I3
(4)
The Kirchhoff Loop Rule going clockwise in the two smaller independent loops gives
V1 − I1 R1 − I2 R2 = 0
(5)
V2 − I3 R3 + I2 R2 = 0
(6)
Add (5) to (6) and solve for I3 to get
I3 =
V1 + V2 − I1 R1
R3
(7)
Substitute (4) into (5) and then substitute (7) into the modified (5). This gives
V1 + V2 − I1 R1
V1 − I1 R1 − I1 −
R2 = 0
R3
which we multiply by R3 and solve for I1 to get
I1 =
V1 (R2 + R3 ) + V2 R2
= 43.3 mA
R1 R2 + R1 R3 + R2 R3
The dissipated power is P1 = I12 R1 = 0.40 W, so the 0.5 W resistor R1 will not burn up.