\text{Chapter 27}

\text{Five resistors are connected as shown in FIGURE 5. The potential difference between }

\text{points $\mathbf{A}$ and $\mathbf{B}$ is $25.0 \mathrm{~V}$. What is the current through the $1.80$ $ \Omega$ resistor?}

I_1

I_2

\xi

A

B

\text{A) 2.76 A}

\text{B) 3.34 A}

\text{C) 1.67 A}

\text{D) 0.577 A}

\text{E) 2.09 A}

\text{Five resistors are connected as shown in FIGURE 5. The potential difference between }

\text{points $\mathbf{A}$ and $\mathbf{B}$ is $25.0 \mathrm{~V}$. What is the current through the $1.80$ $ \Omega$ resistor?}

I=\frac{25.0}{3.60+2.40+1.49}=3.34 \mathrm{~A}

\begin{aligned} & \xi-3.60 \mathrm{I}-1.80 \mathrm{I}_1-2.40 \mathrm{I}=0 \\ & \Rightarrow \mathrm{I}_1=\frac{\varepsilon-6 \mathrm{I}}{1.8}=\frac{25-(6 \times 3.34)}{1.8}=2.76 \mathrm{~A} \end{aligned}

\text{Now, consider loop 1:}\\

\text{(We used the equivalent resistance)}

\text{Answer A}

I_1

I_2

\xi

A

B

\xi

A

B

1.49 \hspace{1mm}\Omega

1

\text{In the circuit shown in FIGURE 6, the current $\boldsymbol{I}=0.36 \mathrm{~A}$.}

\text{What is the potential difference $\boldsymbol{V}_A-\boldsymbol{V}_B$ ?}

I

\cdot

\cdot

A

B

\text{A) $+$2.9 V}

\text{B) $+$1.1 V}

\text{C) $-$1.1 V}

\text{D) $-$2.9 V}

\text{E) $-$4.7 V}

\text{In the circuit shown in FIGURE 6, the current $\boldsymbol{I}=0.36 \mathrm{~A}$.}

\text{What is the potential difference $\boldsymbol{V}_A-\boldsymbol{V}_B$ ?}

\begin{aligned} & +8.0-3.0 \mathrm{I}-5.0 \mathrm{I}_{\mathrm{x}}=0 \\ & \Rightarrow \mathrm{I}_{\mathrm{x}}=\frac{8.0-3.0 \mathrm{I}}{5.0}=\frac{8.0-1.08}{5.0}=1.384 \mathrm{~A} \end{aligned}

\begin{aligned} & \mathrm{V}_{\mathrm{A}}+4.0-5.0 \mathrm{I}_{\mathrm{x}}=\mathrm{V}_{\mathrm{B}} \\ & \Rightarrow \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=5.0 \mathrm{I}_{\mathrm{x}}-4.0=+2.9 \mathrm{~V} \end{aligned}

\text{Now, proceed from $\mathrm{A} \rightarrow \mathrm{B}$ through the $4.0 \mathrm{~V}$ battery:}

\text{Consider Loop 1:}

\text{Answer A}

I

\cdot

\cdot

A

B

1

\text { Find the value of the } \operatorname{emf}(\mathrm{E}) \text { of the battery shown in Figure 6, if the current } \mathrm{I}=1.2 \mathrm{~A} \text {. }

\text{A) 1.8 V}

\text{B) 7.5 V}

\text{C) 3.8 V}

\text{D) 3.2 V}

\text{E) 4.3 V}

\text { Find the value of the } \operatorname{emf}(\mathrm{E}) \text { of the battery shown in Figure 6, if the current } \mathrm{I}=1.2 \mathrm{~A} \text {. }

\text{loop 3}

\begin{aligned} & 15-5 \times 1.2+10 i_2-15=0 \Rightarrow i_2=0.6 \mathrm{~A} \\ & i_3=i_1+i_2=1.8 \mathrm{~A} \end{aligned}

\underline{\text{loop 3}}

\begin{aligned} & 15-5 \times 1.2-4 \times 1.8-\xi=0 \\ & \xi=1.8 \mathrm{~V} \end{aligned}

\underline{\text{loop 1}}

i_1

i_2

i_3

\text{Answer A}

1

\text{In the circuit shown in Figure 7, what should be the ratio $\xi_3 / \xi_1$ if $\xi_1=\xi_2$ and the electric}

\text{current in the circuit equal to zero?}

+

+

+

-

-

-

R

R

R

\text{A) 2.0}

\text{B) 0.5}

\text{C) 1.0}

\text{D) 4.0}

\text{E) 0.25}

\text{In the circuit shown in Figure 7, what should be the ratio $\xi_3 / \xi_1$ if $\xi_1=\xi_2$ and the electric}

\text{current in the circuit equal to zero?}

+

+

+

-

-

-

R

R

R

\begin{aligned} & \mathrm{i}=0 \Rightarrow \xi_1-\xi_3+\xi_2=0 \\ & \xi_1=\xi_2 \Rightarrow 2 \xi_1-\xi_3=0 \\ & \frac{\xi_3}{\xi_1}=2 \end{aligned}

\text{Answer A}

\text { Consider the circuit shown in FIGURE 6. Find the potential difference } V_a-V_b \text {. }

\cdot

\cdot

a

b

\text{A) $-5.68 \hspace{1mm}V$}

\text{B) $+5.68\hspace{1mm} V$}

\text{C) $+44.3 \hspace{1mm}V$}

\text{D) $+19.3 \hspace{1mm}V$}

\text{E) $-19.3\hspace{1mm} V$}

\text { Consider the circuit shown in FIGURE 6. Find the potential difference } V_a-V_b \text {. }

\frac{1}{\mathrm{R}_{\mathrm{eq}}^*}=\frac{1}{25}+\frac{1}{5}+\frac{1}{10}=\frac{17}{50}

{\mathrm{R}_{\mathrm{eq}}^*}=\frac{50}{17}\hspace{1mm} \Omega=2.94 \text{ }\Omega

\Rightarrow

R_\text{eq}=10+2.94=12.94\text{ }\Omega

\mathrm{i}_{\text {Battery }}=\frac{25}{\mathrm{R}_{\mathrm{eq}}}=1.93 A

\begin{aligned} & \mathrm{V}_{\mathrm{a}}-10 i_{\mathrm{B}}+25=\mathrm{V}_{\mathrm{b}} \\ & \mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}=10 i_{\mathrm{B}}-25=-5.68 \mathrm{~V} \end{aligned}

\text{Answer A}

\cdot

\cdot

a

b

i_B

\text { For the circuit shown in FIGURE 7, what is the current in the upper } 20.0 \Omega \text { resistor? }

5.00 \text{ }\Omega

15.0 \text{ }\Omega

20.0 \text{ }\Omega

10.0 \text{ }\Omega

20.0 \text{ }\Omega

45.0 \text{ }V

10.0\text{ } \Omega

\text{A) 0.571 A}

\text{B) 0.643 A}

\text{C) 0.429 A}

\text{D) 1.290 A}

\text{E) 0.321 A}

\text { For the circuit shown in FIGURE 7, what is the current in the upper } 20.0 \Omega \text { resistor? }

5.00 \text{ }\Omega

15.0 \text{ }\Omega

20.0 \text{ }\Omega

10.0 \text{ }\Omega

20.0 \text{ }\Omega

45.0 \text{ }V

10.0\text{ } \Omega

\frac{1}{R_p}=\frac{1}{20}+\frac{1}{10}+\frac{1}{20}=\frac{1+2+1}{20}=\frac{4}{20} \Rightarrow R_p=5 \Omega

R_{e q}=5+15+5+10=35 \Omega

\mathrm{i}_{\text {Battery }}=\frac{45}{35}=\frac{9}{7} \mathrm{~A} ;

i_\text{Battery}

i

V=R_p i_\text{Battery}=5\times 9/7=6.43 \text{ V}

i=\frac{V}{20}=\frac{6.43}{20}=0.321\text{ A}

\text{The 3 resistors in parallel have the ssame $V$ as $R_p$:}

\text{The current through the 20 $ \Omega$ resistor is:}

\text{Answer E}

\text{In Figure 6, find the equivalent resistance between points a and b}

A) 7.0\hspace{1mm} \Omega\\ B) 10 \hspace{1mm}\Omega\\ C)45 \hspace{1mm}\Omega\\ D) 40 \hspace{1mm}\Omega \\ E) 17 \hspace{1mm}\Omega

\text{Answer:}

\text{where did go the $10$ $\Omega$ resistor?}

r_1=10\hspace{1mm}\Omega

r_2=0\hspace{1mm}\Omega

\frac{1}{r_\text{eq}}=\frac{1}{r_1}+\frac{1}{r_2}=\frac{1}{r_1}+\frac{1}{0}=\frac{1}{r_1}+\infty=\infty

r_\text{eq}=\frac{1}{\infty}=0

\text{the equivalent resistance =0, }\\ \text{which means it is just a wire!}

\text{In Figure 6, find the equivalent resistance between points a and b}

\text{For the circuit given in FIGURE 5, if the current through one of the $6.00 \Omega$ is $0.500 \mathrm{~A}$ find }

\text{the current through the $4.00 \Omega$ resistor.}

6.00 \text{ V}

i_2

\text{For the circuit given in FIGURE 5, if the current through one of the $6.00 \Omega$ is $0.500 \mathrm{~A}$ find }

\text{the current through the $4.00 \Omega$ resistor.}

\text{since we have the value of $i_1$,}

\text{one loop will be sufficient}

1)\text{choose the directions of $i_1$ and $i_2$}

2)\text{choose the direction of the loop (counter-clockwise)}

3)\text{draw the arrows for the potential increase}

a)\text{at emf: \textcolor{green}{towards the +}}

b)\text{at resistors: \textcolor{green}{opposit to current}}

4)\text{the sum of potentials =0: if with the chosen direction it is + if against it is $-$}

6.00 \text{ V}

i_2

\circlearrowleft

\xi

R_1 i_1

R_2 i_2

\text{An emf source with $\mathscr{E}=100 \mathrm{~V}$, a resistor with resistance $R=90.0 \Omega$, and a capacitor with}

\text{capacitance $C=5.00 \mu \mathrm{F}$ are connected in series. As the capacitor charges, what is the charge}

\text{on the capacitor when the current in the resistor is $0.800 \mathrm{~A}$ ?}

\text{An emf source with $\mathscr{E}=100 \mathrm{~V}$, a resistor with resistance $R=90.0 \Omega$, and a capacitor with}

\text{capacitance $C=5.00 \mu \mathrm{F}$ are connected in series. As the capacitor charges, what is the charge}

\text{on the capacitor when the current in the resistor is $0.800 \mathrm{~A}$ ?}

\begin{aligned} & q=\mathrm{q}_m\left(1-e^{-\frac{t}{\tau}}\right)=\mathrm{q}_m-\mathrm{q}_m e^{-\frac{t}{\tau}} \\ & \begin{aligned} i= & \frac{q_m}{\tau} e^{-\frac{t}{\tau}} \Rightarrow e^{-\frac{t}{\tau}}=\frac{i \tau}{q_m} \\ \Rightarrow q & =q_m\left(1-\frac{i \tau}{q_m}\right)=q_m-i \tau \\ & =\mathrm{C} \varepsilon-i \mathrm{RC}=\mathrm{C}(\varepsilon-i \mathrm{R}) \\ & =5 \times(100-72)=140 \mu C \end{aligned} \end{aligned}

\text{Answer D}

\text{A $1.0 \mu \mathrm{F}$ capacitor with an initial stored energy of 0.50 J is discharged through 1.0 $\mathrm{M} \Omega$ resistor.}

\text{Find the charge on the capacitor at $\mathrm{t}=0.40 \mathrm{~s}$.}

\text{A) $6.7 \times 10^{-4} \mathrm{C}$}

\text{B) $3.7 \times 10^{-4} \mathrm{C}$}

\text{C) $1.3 \times 10^{-4} \mathrm{C}$}

\text{D) $9.4 \times 10^{-4} \mathrm{C}$}

\text{E) $7.3 \times 10^{-4} \mathrm{C}$}

\begin{aligned} & \mathrm{Q}=\mathrm{Q}_0 \mathrm{e}^{-\mathrm{t} / \mathrm{RC}} \\ & \mathrm{U}_0=\frac{1}{2} \frac{\mathrm{Q}_0{ }^2}{\mathrm{C}} \Rightarrow \mathrm{Q}_0=\sqrt{2 \mathrm{CU}_0} \\ & \mathrm{Q}=\sqrt{2 \mathrm{CU}_0} \mathrm{e}^{-\mathrm{t} / \mathrm{RC}}=6.7 \times 10^{-4} \mathrm{C} \end{aligned}

\text{A $1.0 \mu \mathrm{F}$ capacitor with an initial stored energy of 0.50 J is discharged through 1.0 $\mathrm{M} \Omega$ resistor.}

\text{Find the charge on the capacitor at $\mathrm{t}=0.40 \mathrm{~s}$.}

\text{A) $6.7 \times 10^{-4} \mathrm{C}$}

\text{B) $3.7 \times 10^{-4} \mathrm{C}$}

\text{C) $1.3 \times 10^{-4} \mathrm{C}$}

\text{D) $9.4 \times 10^{-4} \mathrm{C}$}

\text{E) $7.3 \times 10^{-4} \mathrm{C}$}

\cdot

\cdot

\cdot

S

\begin{aligned}&\text{A)2.3A}\\&\text{B)1.6A}\\&\text{C)28mA}\\&\text{D)68mA}\\&\text{E)97mA}\end{aligned}

\begin{aligned} &\text{In the circuit shown in Figure 8, resistances $R_{1}=15\hspace{1mm}\Omega,R_{2}=20\hspace{1mm}\Omega$, and $R_{3}=60\hspace{1mm}\Omega.$ When }\\ &\text{switch S is open, the current in $R_1$ is 2.0 A. What is the current in $R_1$ when the switch is closed?} \end{aligned}

C

R_1

R_2

\xi

s

A) 1.50 \hspace{1mm}mA\\ B) 0.75 \hspace{1mm}mA\\ C) 3.00 \hspace{1mm}mA\\ D) 4.10\hspace{1mm}mA\\ \hspace{-6mm} E) Zero

\begin{aligned} &\text{In the circuit shown in Figure 9, $R_1=R_2=3.0\hspace{1mm}k\Omega$, $C=12\hspace{1mm}\mu F$ and emf $\xi=4.5 \hspace{1mm}V$}\\ & \text{Switch S is closed at t=0. After a long time, what is the current through resistor $R_2$}? \end{aligned}

\text{A segment of a circuit diagram is shown in Figure 6. At a particular instant, if $R=2.0 \mathrm{k} \Omega$,}

\text{$C=4.0 \mathrm{mF}, \varepsilon=8.0 \mathrm{~V}, Q=20 \mathrm{mC}$, and $i=3.0 \mathrm{~mA}$, what is the potential difference $V_a-V_b$ ?}

\text{A segment of a circuit diagram is shown in Figure 6. At a particular instant, if $R=2.0 \mathrm{k} \Omega$,}

\text{$C=4.0 \mathrm{mF}, \varepsilon=8.0 \mathrm{~V}, Q=20 \mathrm{mC}$, and $i=3.0 \mathrm{~mA}$, what is the potential difference $V_a-V_b$ ?}

\text{Answer A}