Problems & Exercises
(a) 0.74 A
(b) 0.742 A
(a) 60.8 W
(b) 3.18 kW
(b) Resistance cannot be negative.
(c) Series resistance is said to be less than one of the resistors, but it must be greater than any of the resistors.
(a) 0.658 A
(b) 0.997 W
(c) 0.997 W; yes
(a) 200 A
(b) 10.0 V
(c) 2.00 kW
(b) No, there is only one independent equation, so only can be found.
(a) –0.120 V
(c) Negative terminal voltage; negative load resistance.
(d) The assumption that such a cell could provide 8.50 A is inconsistent with its internal resistance.
(a) No, you would get inconsistent equations to solve.
(b) . The assumed currents violate the junction rule.
(a) 1.58 mA
(b) 1.5848 V (need four digits to see the difference)
(c) 0.99990 (need five digits to see the difference from unity)
(c) 0.9980, or a percent decrease
(d) 1.002, or a percent increase
(e) Not significant.
(b) You can’t have negative resistance.
(c) It is unreasonable that is greater than (see Figure 21.33). You cannot achieve a full-scale deflection using a current less than the sensitivity of the galvanometer.
(a) 2.00 V
(b) 2.00 s
(b) 30.0 ms
(a) 20.0 s
(b) 120 s
(c) 16.0 ms
(a) 4.99 s
Test Prep for AP® Courses
(a) 4-Ω resistor; (b) combination of 20-Ω, 20-Ω, and 10-Ω resistors; (c) 20 W in each 20-Ω resistor, 40 W in 10-Ω resistor, 64 W in 4-Ω resistor, total 144W total in resistors, output power is 144 W, yes they are equal (law of conservation of energy); (d) 4 Ω and 3 Ω for part (a) and no change for part (b); (e) no effect, it will remain the same.
0.25 Ω, 0.50 Ω, no change
- I1 + I3 = I2
- E1 - I1R1 - I2R2 - I1r1 = 0; - E2 + I1R1 - I3R3 - I3r2 = 0
- I1 = 8/15 A, I2 = 7/15 A and I3 = -1/15 A
- I1 = 2/5 A, I2 = 3/5 A and I3 = 1/5 A
- PE1 = 18/5 W and PR1 = 24/25 W, PR2 = 54/25 W, PR3 = 12/25 W. Yes, PE1 = PR1+ PR2 + PR3
- R3, losses in the circuit
(a) 20 mA, Figure 21.41, 5.5 s; (b) 24 mA, Figure 21.32, 2 s