POWER IN ELECTRIC CIRCUITS

𝑃 = 𝐼𝑉 𝑃 𝛼 𝐼, 𝑉 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝑒𝑠𝑝𝑒𝑐𝑖𝑎𝑙𝑙𝑦 𝑖𝑛 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑐𝑖𝑟𝑐𝑢𝑖𝑡𝑠

ℎ𝑒𝑛𝑐𝑒: 𝑡ℎ𝑒 𝑙𝑜𝑤𝑒𝑟 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟 𝒃𝒖𝒓𝒏𝒔 𝒃𝒓𝒊𝒈𝒉𝒕𝒆𝒓

P = 𝐼2𝑅 𝑃 𝛼 𝑅, 𝐼 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑖𝑛 𝑠𝑒𝑟𝑖𝑒𝑠 𝑐𝑖𝑟𝑐𝑢𝑖𝑡 ,

HENCE, LOWER RESISTOR BURNS DIM

𝑃 =𝑉2

𝑅(𝑃 𝛼 𝑉2), 𝑅 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

PROBLEM 65 CHAPTER 20Information to use• When two or more resistors are in series, the equivalent resistance is given by

Rs R1 R2 R3 . . .

• When resistors are in parallel, the expression to be solved to find the equivalent resistance is given by

1

Rp

1

R1

1

R2

1

R3

....

We will successively apply these to the individual resistors in the figure in the text

beginning with the resistors on the right side of the figure.

• Since the 4.0- and the 6.0- resistors are in series, the equivalent resistance of the combination of those two resistors is 10.0 .

• The 9.0- and 8.0- resistors are in parallel; their equivalent resistance is 4.24 .

• The equivalent resistances of the parallel combination (9.0 and 8.0 ) and the series combination (4.0 and the 6.0 ) are in parallel; therefore,

equivalent resistance is 2.98 .

• The 2.98- combination is in series with the 3.0- resistor, so that equivalent resistance is 5.98 .

• Finally, the 5.98- combination and the 20.0- resistor are in parallel, so the equivalent resistance between the points A and B is4 6.

.

Related problem no 64 page 632

PROBLEM 85 CHAPTER 20STEP 1: CHOOSE A JUNCTION AND DIRECTIONS OF CURRENTS AND ALLOCATE SIGNS ON RESISTORS

STEP 2: APPLY THE JUNCTION RULE :

𝑰𝟏 + 𝑰𝟑 = 𝑰𝟐………………(𝟏)

STEP 3: APPLYING THE LOOP RULE

LOOP ABCD CLOCKWISE:

−𝟐 𝑰𝟏 + 𝟔𝑽 + 𝟒 𝑰𝟑 + 𝟑𝑽 = 𝟎 …………… 𝟐

LOOP BEFC CLOCKWISE:

−𝟖 𝑰𝟐 − 𝟗 𝑽− 𝟒 𝑰𝟑 − 𝟔𝑽 = 𝟎 …………… 𝟑

SOLVING SIMULTANEOUS EQUATIONS WE GET

𝑰𝟑 = −𝟏. 𝟖𝟐 𝑨

SIGNIFICANCE OF MINUS SIGN:

B

C

6.00 V

A

D

E

F

2.00 Ω

I1 I2

8.00 Ω

I3 4.00 Ω

9.00 V3.00 V+ + +

+ +

+

The minus sign indicates that the current in the 4.00- resistor is directed downward , rather

than upward as selected arbitrarily in the drawing.

CHAPTER 7: PROBLEM 25

REASONING• No net external force acts on the plate parallel to the floor; therefore, the

component of the momentum of the plate that is parallel to the floor is conserved as the plate breaks and flies apart.

• Initially, the total momentum parallel to the floor is zero. After the collision with the floor, the component of the total momentum parallel to the floor must remain zero.

• The drawing in the text shows the pieces in the plane parallel to the floor just after the collision.

• Clearly, the linear momentum in the plane parallel to the floor has two components; therefore the linear momentum of the plate must be conserved in each of these two mutually perpendicular directions.

• Using the drawing in the text, with the positive directions taken to be up and to the right, we have

m v m v1 1 2 2

(sin 25.0 ) + (cos 45.0 ) = 0

(1)

y direction m v m v m v1 1 2 2 3 3

(cos 25.0 ) + (sin 45.0 ) – = 0

(2)

m1 1.00 kg

m2 1.00 kg

Substituting known values and solving equations simultaneously we get

.Related example 9 section 7.4 in your prescribed book: Cutnell and Johnsons 9th Edition