• Home
  • Textbooks
  • Schaum’s Outline of College Physics
  • Kirchhoff's Laws

Schaum’s Outline of College Physics

Eugene Hecht

Chapter 29

Kirchhoff's Laws - all with Video Answers

Educators

Chapter Questions

07:16

Problem 1

Find the currents in the circuit shown in Fig. $29-1$.
Notice that the signs of the voltage drops have been provided in the circuit diagram. You will not need them in this solution, but it's a good habit to put them in as a first step.
This circuit cannot be reduced further because it contains no resistors in simple series or parallel combinations. We therefore revert to Kirchhoff's rules. If the currents had not been labeled and shown by arrows, we would do that first. In general, special care is needed in assigning the current directions, since those chosen incorrectly will simply give negative numerical values. In this problem there are three branches connecting nodes- $a$ and $-b$, and therefore three currents. Apply the node rule to node- $b$ in Fig. $29-1$ :
$$
\begin{array}{l}
\text { Current into } b=\text { Current out of } b\\
I_{1}+I_{2}+I_{3}=0
\end{array}
$$
Next apply the loop rule to loop adba. In volts,
$$
-7.0 I_{1}+6.0+4.0=0 \quad \text { or } \quad I_{1}=\frac{10.0}{7.0} \mathrm{~A}
$$
(Why must the term $7.0 I_{1}$ have a negative sign?) Then apply the loop rule to loop $a b c a$. In volts,
$$
-4.0-8.0+5.0 I_{2}=0 \quad \text { or } \quad I_{2}=\frac{12.0}{5.0} \mathrm{~A}
$$
(Why must the signs be as written?)
Now return to Eq. (1) to find
$$
I_{3}=-I_{1}-I_{2}=-\frac{10.0}{7.0}-\frac{12.0}{5.0}=\frac{-50-84}{35}=-3.8 \mathrm{~A}
$$
The minus sign tells us that $I_{3}$ is opposite in direction to that shown in the figure.

Sikandar Baig
Sikandar Baig
Numerade Educator
10:20

Problem 2

For the circuit shown in Fig. 29-2, find $I_{1}, I_{2}$, and $I_{3}$ if switch $S$ is $(a)$ open and $(b)$ closed.
(a) When $S$ is open, $I_{3}=0$, because no current can flow through the middle branch. Applying the node rule to point $-a$
$$
I_{1}+I_{3}=I_{2} \quad \text { or } \quad I_{2}=I_{1}+0=I_{1}
$$
Applying the loop rule to the outer loop acbda yields
$$
-12.0+7.0 I_{1}+8.0 I_{2}+9.0=0
$$
To understand the use of signs, remember that current always flows from high to low potential through
a resistor.
Because $I_{2}=I_{1}$, Eq. (1) becomes
$$
\begin{array}{lll}
15.0 I_{1}=3.0 & \text { or } & I_{1}=0.20 \mathrm{~A}
\end{array}
$$
Also, $I_{2}=I_{1}=0.20 \mathrm{~A}$. Notice that this is the same result that one would obtain by replacing the two batteries by a single $3.0$ - $\mathrm{V}$ battery.
(b) With $S$ closed, $I_{3}$ is no longer necessarily zero. Applying the node rule to point- $a$ gives
$$
I_{1}+I_{3}=I_{2}
$$
Applying the loop rule to loop acba
$$
-12.0+7.01 I_{1}-4.0 I_{3}=0
$$
and to loop adba gives
$$
-9.0-8.0 I_{2}-4.0 I_{3}=0
$$
Applying the loop rule to the remaining loop, acbda, would yield a redundant equation, because it would contain no new voltage change.
Now solve Eqs. (2), (3), and (4) for $I_{1}, I_{2}$, and $I_{3}$. From Eq. (4),
$$
I_{3}=-2.0 I_{2}-2.25
$$
Substituting this in Eq. (3) yields
$$
-12.0+7.0 I_{1}+9.0+8.0 I_{2}=0 \quad \text { or } \quad 7.0 I_{1}+8.0 I_{2}=3.0
$$
Substituting for $I_{3}$ in Eq. (2) also gives
$$
\begin{array}{lll}
I_{1}-2.0 I_{2}-2.25-I_{2} & \text { or } & I_{1}=3.0 I_{2}+2.25
\end{array}
$$
Substituting this value in the previous equation finally leads to
$$
21.0 I_{2}+15.75+8.0 I_{2}=3.0 \quad \text { or } \quad I_{2}=-0.44 \mathrm{~A}
$$
Using this in the equation for $I_{1}$,
$$
I_{1}=3.0(-0.44)+2.25=-1.32+2.25=0.93 \mathrm{~A}
$$
Notice that the minus sign is a part of the value we have found for $I_{2}$. It must be carried along with its numerical value. Now use (2) to find
$$
I_{3}=I_{2}-I_{1}=(-0.44)-0.93=-1.37 \mathrm{~A}
$$

Sikandar Baig
Sikandar Baig
Numerade Educator
10:52

Problem 3

Each of the cells shown in Fig. $29-3$ has an emf of $1.50 \mathrm{~V}$ and a $0.0750-\Omega$ internal resistance. Find $I_{1}, I_{2}$, and $I_{3}$.
Applying the node rule to point- $a$ gives
$$
I_{1}=I_{2}+I_{3}
$$
Applying the loop rule to loop abcea yields, in volts,
$$
-(0.0750) I_{2}+1.50-(0.0750) I_{2}+1.50-3.00 I_{1}=0
$$
or
$$
3.00 I_{1}+0.150 I_{2}=3.00
$$
Also, for loop adcea,
or
$$
\begin{array}{r}
-(0.0750) I_{3}+1.50-(0.0750) I_{3}+1.50-3.00 I_{1}=0 \\
3.00 I_{1}+0.150 I_{3}=3.00
\end{array}
$$
Solve Eq. (2) for $3.00 I_{1}$ and substitute in Eq. (3) to get
$$
\begin{array}{ccc}
3.00-0.150 I_{3}+0.150 I_{2}=3.00 & \text { or } & I_{2}=I_{3}
\end{array}
$$
as we might have guessed from the symmetry of the problem. Then Eq. (1) yields
$$
I_{1}=2 I_{2}
$$
and substituting this in Eq. (2),
$$
\begin{array}{lll}
6.00 I_{2}+0.150 I_{2}=3.00 & \text { or } & I_{2}=0.488 \mathrm{~A}
\end{array}
$$
Then, $I_{3}=I_{2}=0.488 \mathrm{~A}$ and $I_{1}=2 I_{2}=0.976 \mathrm{~A}$.

Sikandar Baig
Sikandar Baig
Numerade Educator
09:56

Problem 4

The currents are steady in the circuit of Fig. 29-4. Find $I_{1}, I_{2}, I_{3}, I_{4}, I_{5}$, and the charge on the capacitor.
The capacitor passes no current when charged, and so $I_{5}=0 .$ Consider loop acba. The loop rule leads to
$$
-8.0+4.0 I_{2}=0 \quad \text { or } \quad I_{2}=2.0 \mathrm{~A}
$$
Using loop adeca gives
$$
-3.0 I_{1}-9.0+8.0=0 \quad \text { or } \quad I_{1}=-0.33 \mathrm{~A}
$$
Applying the node rule at point- $c$ results in
$$
I_{1}+I_{5}+I_{2}=I_{3} \quad \text { or } \quad I_{3}=1.67 \mathrm{~A}=1.7 \mathrm{~A}
$$
and at point- $a$, it yields
$$
I_{3}=I_{4}+I_{2} \quad \text { or } \quad I_{4}=-0.33 \mathrm{~A}
$$
(We should have realized this at once, because $I_{5}=0$ and so $I_{4}=I_{1} .$.)
To find the charge on the capacitor, we need the voltage $V_{f g}$ across it. Put in all the signs on the resistors, batteries, and capacitor. Applying the loop rule to loop dfgced gives
$$
-2.0 I_{5}+V_{f g}-7.0+9.0+3.0 I_{1}=0 \quad \text { or } \quad 0+V_{f g}-7.0+9.0-1.0=0
$$
from which $V_{f g}=-1.0 \mathrm{~V}$. The minus sign tells us that plate $g$ is negative. The capacitor's charge is
$$
Q=C V=(5.0 \mu \mathrm{F})(1.0 \mathrm{~V})=5.0 \mu \mathrm{C}
$$

Vishal Gupta
Vishal Gupta
Numerade Educator
09:49

Problem 5

For the circuit shown in Fig. $29-5$, the resistance $R$ is $5.0 \Omega$ and $\mathscr{E}=20 \mathrm{~V}$. Find the readings of the ammeter and the voltmeter. Assume the meters to be ideal.
The ideal voltmeter has infinite resistance (no current passes through it), and so it can be removed from the circuit with no effect. Write the loop equation for loop cdefc:
$$
-R I_{1}+12.0-8.0-7.0 I_{2}=0
$$
which becomes
$$
5.0 I_{1}+7.0 I_{2}=4.0
$$
Next write the loop equation for loop cdeac. It is
$$
\begin{array}{c}
-5.0 I_{1}+12.0+2.0 I_{3}+20.0=0 \\
5.0 I_{1}-2.0 I_{3}=32.0
\end{array}
$$
But the node rule applied at $e$ gives
$$
I_{1}+I_{3}=I_{2}
$$
Substituting Eq. (3) in Eq. (1) yields
$$
5.0 I_{1}+7.0 I_{1}+7.0 I_{3}=4.0
$$
Solve this for $I_{3}$ and substitute in (2) to get
$$
5.0 I_{1}-2.0\left(\frac{4.0-12.0 I_{1}}{7.0}\right)=32.0
$$
which yields $I_{1}=3.9 \mathrm{~A}$, which is the ammeter reading. Then Eq. (1) gives $I_{2}=-2.2 \mathrm{~A}$.
To find the voltmeter reading $V_{a b}$, write the loop equation for loop $a b c a:$
$$
V_{a b}-7.0 I_{2}-\mathscr{E}=0
$$
Substituting the known values of $I_{2}$ and $\mathscr{E}$, then solving, we obtain $V_{a b}=4.3 \mathrm{~V}$. Since this is the potential difference between $a$ to $b$, point $b$ must be at the higher potential.

Sikandar Baig
Sikandar Baig
Numerade Educator
07:24

Problem 6

In the circuit in Fig. $29-5, I_{1}=0.20 \mathrm{~A}$ and $R=5.0 \Omega .$ Find $\mathscr{E}$.
We write the loop equation for loop cdefc:
$$
-R I_{1}+12.0-8.0-7.0 I_{2}=0 \quad \text { or } \quad-(5.0)(0.20)+12.0-8.0-7.0 I_{2}=0
$$
from which $I_{2}=0.43$ A. We can now find $I_{3}$ by applying the node rule at $e$ :
$$
I_{1}+I_{3}=I_{2} \quad \text { or } \quad I_{3}=I_{2}-I_{1}=0.23 \mathrm{~A}
$$
Now apply the loop rule to loop cdeac:
$$
-(5.0)(0.20)+12.0+(2.0)(0.23)+\mathscr{E}=0
$$
from which $\mathscr{E}=-11.5 \mathrm{~V}$. The minus sign tells us that the polarity of the battery is actually the reverse of that shown.

Sikandar Baig
Sikandar Baig
Numerade Educator
12:00

Problem 7

For the circuit shown in Fig. $29-6$, find the current in the $0.96-\Omega$ resistor and the terminal voltages of the batteries.

DM
Debra Mangion
Numerade Educator
09:48

Problem 8

For the network shown in Fig. $29-7$, determine $(a)$ the three currents $I_{1}, I_{2}$, and $I_{3}$, and $(b)$ the terminal voltages of the three batteries.

Sikandar Baig
Sikandar Baig
Numerade Educator
01:22

Problem 9

Refer back to Fig. $29-5 .$ If the voltmeter reads $16.0 \mathrm{~V}$ (with point- $b$ at the higher potential) and $I_{2}=0.20 \mathrm{~A}$, find $\mathscr{E}, R$, and the ammeter reading.

Anand Jangid
Anand Jangid
Numerade Educator
08:12

Problem 10

Find $I_{1}, I_{2}, I_{3}$, and the potential difference between point- $b$ to point- $e$ in Fig. $29-8$.

Sikandar Baig
Sikandar Baig
Numerade Educator
08:54

Problem 11

In Fig. $29-9, R=10.0 \Omega$ and $\mathscr{E}=13 \mathrm{~V}$. Find the readings of the ideal ammeter and voltmeter.

Sikandar Baig
Sikandar Baig
Numerade Educator
01:22

Problem 12

In Fig. 29-9, the voltmeter reads $14 \mathrm{~V}$ (with point- $a$ at the higher potential) and the ammeter reads $4.5 \mathrm{~A}$. Find $\mathscr{E}$ and $R$.

Anand Jangid
Anand Jangid
Numerade Educator