Presentation is loading. Please wait.

Presentation is loading. Please wait.

(b) ρ L R = A 1.59 x 10-8 (1.8) (a) (c) (d) (1) GIVEN = A

Similar presentations


Presentation on theme: "(b) ρ L R = A 1.59 x 10-8 (1.8) (a) (c) (d) (1) GIVEN = A"— Presentation transcript:

1 (b) ρ L R = A 1.59 x 10-8 (1.8) (a) (c) (d) (1) GIVEN 0.993 = A
V = 4.50 V L = 180 cm = 1.8 m ρ = 1.59 x 10-8 Ωm I = 4.53 A (b) ρ L R = A 1.59 x 10-8 (1.8) 0.993 = A A = 2.9 x 10-8 m2 (a) V = IR (c) A = πr2 4.5 = 4.53(R) 2.9 x 10-8 = πr2 R = Ω r = 9.6 x 10-5 m d = 1.92 x 10-4 m (d) Don’t start all over; 45 cm is ¼ of 180 cm ∴ R will ¼ of 0.99 Ω or Ω

2 (2) VT = IT RT 120 = I (21.4) RT = 5 + 1.2 + 10 + 4 + 1.2 = 21.4 Ω
3 Ω 10 Ω 1.2 2 Ω 4 Ω 120 V 4 Ω 2 Ω 6 Ω 12 1.2 6 Ω VT = IT RT 120 = I (21.4) RT = = 21.4 Ω IT = 5.6 amps This is the current for each single resistor and each group

3 10Ω I = 5.6 amps V = 28 V P = 157 watts I = 5.6 amps V = 56 V P = watts I = 5.6 amps V = 22.4 V P = watts

4 V= I R V = 5.6(1.2) V = 6.7 V V3= I3R3 6.7 = I3(3) I3 = 2.23 amps
3 Ω V = 5.6(1.2) 1.2 V = 6.7 V 2 Ω This is the voltage for each resistor V3= I3R3 6.7 = I3(3) I3 = 2.23 amps V2= I2R2 6.7 = I2(2) I2 = 3.35 amps

5 10Ω I = 5.6 amps V = 28 V P = 157 watts I = 2.23 amps V = 6.7 V P = 14.9 watts I = 3.35 amps V = 6.7 V P = 22.4 watts I = 5.6 amps V = 56 V P = watts I = 5.6 amps V = 22.4 V P = watts

6 V6= I6R6 V6 = (.56)(6) V6 = 3.36 V V= I R V = 5.6(1.2) V = 6.7 V
4 Ω V6 = 3.36 V 2 Ω 6 Ω 12 1.2 6 Ω V= I R V = 5.6(1.2) This is voltage for the 2Ω and 4Ω resistors and the 12Ω group V = 6.7 V V4= I4R4 6.7 = I3(4) V= I R 6.7 = I(12) I4 = 1.68 amps This is the current for each 6Ω resistor I = 0.56 amps V2= I2R2 6.7 = I2(2) I2 = 3.35 amps

7 10Ω I = 5.6 amps V = 28 V P = 157 watts I = 2.23 amps V = 6.7 V P = 14.9 watts I = 3.35 amps V = 6.7 V P = 22.4 watts I = 5.6 amps V = 56 V P = watts I = 5.6 amps V = 22.4 V P = watts I = 0.56 amps V = 3.36 V P = 1.9 watts I = 0.56 amps V = 3.36 V P = 1.9 watts I = 3.35 amps V = 6.7 V P = 22.4 watts I = 1.68 amps V = 6.7 V P = 11.3 watts

8 (a) (3) 1 1 1 = + RT 14 22 RT = 8.55 Ω VT = IT RT 25 = I (8.55)
Before switch is closed = + RT 14 22 RT = 8.55 Ω 12 Ω 4 Ω VT = IT RT 14 22 25 = I (8.55) 25 V 10 Ω 10 Ω IT = 2.92 amps R1 and R2 are in series; R3 and R4 are in series Both series groups are in parallel V14= I14R14 Each parallel group gets a voltage of 25 V V22= I22R22 25 = I14(14) 25 = I22(22) I14 = 1.78 amps I22 = 1.14 amps R1 and R2 R3 and R4

9 (b) 1 1 1 After switch is closed = + R 4 12 R = 3 Ω 12 Ω 4 Ω 3 1 1 1 = + 10 Ω 10 Ω R 10 10 5 R = 5 Ω R1 and R3 are in parallel; R2 and R4 are in parallel Both parallel groups are in series RT = 8 Ω VT = IT RT Each group gets a current of amps 25 = I (8) IT = amps

10 (b) V= IR V= IR V = (3.125)(3) V = (3.125)(5) V = 9.375 V V = 15.625 V
Solve for the I for each resistor V= IR V= IR V = (3.125)(3) V = (3.125)(5) V = V V = V This is the voltage for R2 and R4 This is the voltage for R1and R3 V= IR V= IR V= IR 9.375 = I(12) = I(10) 9.375 = I(4) I = amps I = 0.78 amps I = 2.34 amps R3 R1 R2 and R4

11 (c) After switch is closed Before switch is closed 7.12 V 13.68 V 4 Ω 12 Ω V 4 Ω 12 Ω V 1.14 amps 1.78 amps 10 Ω 10 Ω 10 Ω 10 Ω Left branch drops 7.12 V while right branch drops V Do the same thing Difference in potential between the left and right branches is 6.56 V V = 0 V after closed

12 (d) I = 2.34 amps through R1 I = 1.56 amps through R2
After switch is closed (d) I = 2.34 amps through R1 and I = 1.56 amps through R2 Thus 2.34 – 1.56 amps flows from a to b 0.781 amps from a to b


Download ppt "(b) ρ L R = A 1.59 x 10-8 (1.8) (a) (c) (d) (1) GIVEN = A"

Similar presentations


Ads by Google