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

(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

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

(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.

(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) $I_1\ne I_2+I_3$. The assumed currents violate the junction rule.

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

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

(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 21.30). 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) $P=\frac{V^2}{R}=\frac{{(1.00\times {10}^2)}^2}{2.50\times {10}^3}\text{w}=4.00\text{ W}$

(b) $\begin{array}{l}\frac{1}{R_{eq}}=\frac{2}{R_1}\\ R_{eq}=\frac{R_1}{2}=1.25\times {10}^3\Omega \\ P=\frac{V^2}{R}=\frac{{(1.00\times {10}^2)}^2}{1.25\times {10}^3}\text{ W}=8\text{W}\end{array}$

(c) $\begin{array}{l}{R}_{eq}=R_1+R_2=2R_1\\ P=\frac{V^2}{R}=\frac{{(1.00\times {10}^2)}^2}{5.00\times {10}^3}\text{W}=2\text{W}\end{array}$

(d) In the parallel case current passes through each branch, so more current is passed through the resistor system. In the series, the higher resistance of the circuit allows less current to flow. $P=IV$ is proportional to the current when the voltage does not change.

(e) The parallel combination delivers more energy in a given time period.

(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 21.37, 5.5 s; (b) 24 mA, Figure 21.29, 2 s