Chapter 21

(a) $2\text{.}\text{75}\text{k}\Omega$

(b) $\text{27}\text{.}5\Omega$

(a) $\text{786}\Omega$

(b) $\text{20}\text{.}3\Omega$

$\text{29}\text{.}6\text{W}$

(a) 0.74 A

(b) 0.742 A

(a) 60.8 W

(b) 3.18 kW

(a) $\begin{array}{}{R}_{\text{s}}=R_1+R_2\\ \Rightarrow R_{\text{s}}\approx R_1\left(R_1\text{>>}{R}_2\right)\end{array}$

(b) $\frac{1}{R_{\mathrm{p}}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{R_1+R_2}{R_1{R}_2}$,

so that

$\begin{array}{}{R}_{\mathrm{p}}=\frac{R_1{R}_2}{R_1+R_2}\approx \frac{R_1{R}_2}{R_1}=R_2\left(R_1\text{>>}{R}_2\right)\text{.}\end{array}$

(a) $-\text{400 k}\Omega$

(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.

2.00 V

2.9994 V

$0\text{.}\text{375}\Omega$

(a) 0.658 A

(b) 0.997 W

(c) 0.997 W; yes

(a) 200 A

(b) 10.0 V

(c) 2.00 kW

(d) $0\text{.}\text{1000}\Omega ;\text{80}\text{.}\text{0 A, 4}\text{.}\text{0 V, 320 W}$

(a) $0\text{.}\text{400 Ω}$

(b) No, there is only one independent equation, so only $r$ can be found.

(a) –0.120 V

(b) $-1\text{.}\text{41}\times {\text{10}}^{-2}\Omega$

(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.

$-I_2{R}_2+{\text{emf}}_1-{\text{I}}_2{r}_1+{\text{I}}_3{R}_3+{\text{I}}_3{r}_2-{\text{emf}}_2=\text{0}$

$I_3={\text{I}}_1+{\text{I}}_2$

${\text{emf}}_2-{\text{I}}_2{r}_2-{\text{I}}_2{R}_2+{\text{I}}_1{R}_5+I_1{r}_1-{\text{emf}}_1+{\text{I}}_1{R}_1=0$

(a) ${\text{I}}_1=\text{4.75 A}$

(b) ${\text{I}}_{\text{2}}=-3\text{.}\text{5 A}$$$

(c) ${\text{I}}_3=8\text{.}\text{25 A}$

(a) No, you would get inconsistent equations to solve.

(b) . The assumed currents violate the junction rule.

$\text{30}\mathrm{\mu A}$

$1\text{.}\text{98 k}\Omega$

$1\text{.}\text{25}\times {\text{10}}^{-4}\Omega$

(a) $3\text{.}\text{00 M}\Omega$

(b) $2\text{.}\text{99 k}\Omega$

(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)

$\text{15}\text{.}\mathrm{0\; \mu A}$

(a)

(b) $10\text{.}\text{02}\Omega$

(c) 0.9980, or a $2.0\times {10}^{\mathrm{–1}}$ percent decrease

(d) 1.002, or a $2.0\times {10}^{\mathrm{–1}}$ percent increase

(e) Not significant.

(a) $-\text{66}\text{.}7\Omega$

(b) You can’t have negative resistance.

(c) It is unreasonable that $I_{\mathrm{G}}$ is greater than $I_{\text{tot}}$ (see Figure 21.33). You cannot achieve a full-scale deflection using a current less than the sensitivity of the galvanometer.

24.0 V

$1\text{.}\text{56 k}\Omega$

(a) 2.00 V

(b) $9\text{.}\text{68}\Omega$

$\text{range 4}\text{.}\text{00 to 30}\text{.}\text{0 M}\Omega$

(a) $2\text{.}\text{50 μF}$

(b) 2.00 s

86.5%

(a) $1\text{.}\text{25 k}\Omega$

(b) 30.0 ms

(a) 20.0 s

(b) 120 s

(c) 16.0 ms

$1\text{.}\text{73}\times {\text{10}}^{-2}\text{s}$

$3\text{.}\text{33}\times {\text{10}}^{-3}\Omega$

(a) 4.99 s

(b) $3\text{.}\text{87ºC}$

(c) $\text{31}\text{.}\text{1 k}\Omega$

(d) No

(a), (b)

(b)

(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

a. (c)

b. (c)

c. (d)

d. (d)

1. I₁ + I₃ = I₂
2. E₁ - I₁R₁ - I₂R₂ - I₁r₁ = 0; - E₂ + I₁R₁ - I₃R₃ - I₃r₂ = 0
3. I₁ = 8/15 A, I₂ = 7/15 A and I₃ = -1/15 A
4. I₁ = 2/5 A, I₂ = 3/5 A and I₃ = 1/5 A
5. P_E1 = 18/5 W and P_R1 = 24/25 W, P_R2 = 54/25 W, P_R3 = 12/25 W. Yes, P_E1 = P_R1+ P_R2 + P_R3
6. R_3, losses in the circuit

(a) 20 mA, Figure 21.41, 5.5 s; (b) 24 mA, Figure 21.32, 2 s