Chapter 7, Electric Fields V Video Solutions, Electromagnetic Fields and Waves: Including Electric Circuits

Problem 1

Figure $7-17$ shows a potentiometer circuit. Show that, when $I=0$,
$$
V_{o}=\frac{R_{2}}{R_{1}+R_{2}} V_{i}
$$
This is a common type of circuit. It serves to measure a voltage, in this case $V_{b}$, without drawing current. In curve plotters the current $I$, after amplification, actuates a motor that displaces the pen and simultaneously moves the tap in the direction that decreases $I .$ The resistances $R_{1}$ and $R_{2}$ act as a potential divider.

Vishal Gupta

Numerade Educator

04:07

Problem 2

Figure 7-18(a) shows a common type of amplifier.
The triangular figure is an operational amplifier whose gain is $-A .$ Such amplifiers have gains of the order of $10^{4}$ to $10^{9}$ and draw a negligible amount of current at their input terminals. The accuracy of the gain of this circuit is limited only by the stability of the ratio $R_{2} / R_{1} .$ The drift in $R_{2} / R_{1}$ due to aging, temperature changes, and so forth is normally smaller than the drift in $A$ by orders of magnitude.
As a first approximation, (a) the operational amplifier draws zero current, and the same current $I$ flows in $R_{1}$ and in $R_{2}$, (b) $A$ is infinite, and the potential at the junction between $R_{1}$ and $R_{2}$ is therefore zero. Then $I \approx V_{i} / R_{1} \approx-V_{o} / R_{2}$, and the gain is about $-R_{2} / R_{1}$.
(a) Find a more accurate expression for the gain $V_{o} / V_{i}$. You can take into account the fact that the gain is not infinite by setting $V_{o}=-A V_{i A}$ in Fig. 7-18(b). The approximation $-R_{2} / R_{1}$ is valid only if $A \gg 1$ and if $A \gg>R_{2} / R_{1} .$
(b) What is the minimum value of $A$ if $R_{1}=1000$ ohms, $R_{2}=2000$ ohms, and the circuit gain must be equal to 2 within $0.1 \%$ ?

Khoobchandra Agrawal

Numerade Educator

02:12

Problem 3

Figure $7-19$ shows a so-called $R / 2 R$ ladder network that serves for both digital-to-analog (D/A) and analog-to-digital (A/D) conversion. See Prob. $7-4$

Show that the resistance between $A$ and ground, called the input resistance, is $2 R$, whatever be the number of stages.

Manish Kumar ( Iit K )

Numerade Educator

05:36

Problem 4

The $R / 2 R$ ladder network of Prob. $7-3$ can convert a binary $t$ number to an analog voltage. The usual convention is that of positive logic, in which $V=0$ means 0 and $V=V^{\prime}$ means 1 , where $V^{\prime}$ is a precisely regulated positive voltage.
Refer to Fig. 7-20. Say $V^{\prime}=1$ volt. The voltmeter connected between $A$ and ground draws essentially zero current. The binary number enters at the terminals $B, C, D, \ldots$ and the analog number appears at $A$. For example, if the number is 4 , or $100, V_{B}=1, V_{C}=0, V_{D}=0$.
Use the result of Prob. $7-3$ to find the value of $V_{A}$ when
(a) $V_{B}=1, V_{C}=0$, and $V_{D}=0$;
(b) $V_{B}=0, V_{C}=1$, and $V_{D}=0$;
(c) $V_{B}=0, V_{C}=0$, and $V_{n}=1$

Bruce Edelman

Numerade Educator

04:10

Problem 5

A source charges a capacitor $C$ through a resistor $R$ to a voltage $V$. Calculate the energy supplied by the source, that dissipated by the resistor, and that stored in the capacitor, after an infinite time.
You should find that the resistor dissipates half the energy and that the capacitor stores the other half.

Manish Kumar ( Iit K )

Numerade Educator

03:59

Problem 6

Figure $7-21$ shows an $R C$ differentiating circuit. The load resistance connected at $V_{o}$ is large compared to $R$.
(a) Show that, if the voltage drop across $R$ is negligible compared to that across $C$, then
$$
V_{o} \approx R C \frac{d V_{i}}{d t}
$$
(b) The input is a square wave. Sketch $V_{o}(t)$.

Khoobchandra Agrawal

Numerade Educator

01:20

Problem 7

The circuit of Fig. 7-21(a) is simple and inexpensive, but $V_{o} \ll V_{t}$. Figure 7-22(a) shows a much superior, but more complex, differentiating circuit. The triangle represents an operational amplifier as in Prob. 7-2. Figure $7-22(b)$ shows the equivalent circuit. Show that
$$
V_{o}=-R C \frac{d V_{i}}{d t},
$$
as long as $A \gg 1$ and $\left|V_{o}\right| / R C \gg\left|d V_{o} / d t\right| / A$. Note that $R C$ can be much larger than unity, so that $V_{o}$ need not be much smaller than $V_{i}$.

Khoobchandra Agrawal

Numerade Educator

View

Problem 8

Figure 7-23 shows an $R C$ integrating circuit. The current through the load connected at $V_{o}$ is negligible compared to $d Q / d t$ in $C$.
(a) Show that as long as the voltage across $C$ is small compared to that across $R$,
$$
V_{o}=\frac{1}{R C} \int_{0}^{t} V_{i} d t
$$
As in Prob. $7-6, V_{o} \ll V_{i}$. We assume that $V_{o}=0$ at $t=0$.
(b) Sketch a curve of the output voltage as a function of time if $V_{i}$ is a square wave.

Lainey Roebuck

Numerade Educator

View

Problem 9

The integrating circuit shown in Fig. 7-24(a) performs integrations without the limitation $V_{o} \ll V_{i}$ that applies to the circuit of Fig. 7-23. The triangle represents an operational amplifier, as in Prob. 7-2. Show that
$$
V_{o}=-\frac{1}{R C} \int_{0}^{t} V_{i} d t
$$
if $A \gg 1$ and if $\left|V_{o}\right| /(R C) \ll A\left|d V_{o} / d t\right|$. Use the equivalent circuit of Fig. $7-24(b)$, and set
$$
\frac{d}{d t} V_{o}=\frac{d}{d t}\left(V_{i A}-\frac{Q}{C}\right)
$$

Lainey Roebuck

Numerade Educator

01:38

Problem 10

Figure $7-25$ shows how the differentiating circuit of Prob. 7-7 can serve to measure a displacement.
Show that, with an alternating voltage at the input, the output voltage is proportional to the spacing $s$ of the parallel-plate capacitor, neglecting edge effects.

Mahmoud Abdelshafy

Numerade Educator

01:26

Problem 11

The potential divider of Prob. $7-1$ is not useful as such at high frequencies for the following reason. There are stray capacitances due to the wiring in parallel with $R_{1}$ and $R_{2}$. If the frequency is high enough, these stray capacitances carry an appreciable current and $V_{o} / V_{i}$ is a function of the frequency.
Show that, with the circuit of Fig. 7-26, the relation of Prob. 7-1 applies if $R_{1} C_{1}=R_{2} C_{2} .$ If the added capacitances are large compared to the stray capacitances, the potential divider is said to be compensated.

Thomas Wells

Numerade Educator

03:44

Problem 12

There exist a wide variety of impedance bridge circuits for measuring the impedance of components. These bridges have now been largely superseded by network analyzers that can perform many sophisticated measurements on complex circuits. Figure 7-27(a) shows such an impedance bridge. If $Z_{1} / Z_{2}=Z_{3} / Z_{4}$, then $V=0$. The impedances must satisfy two independent equations to satisfy both the real and the imaginary parts of this equation.
One common type is the Wien bridge shown in Fig. 7-27(b). As a rule, one sets $R_{1}=R_{2} / 2, R_{3}=R_{4}, C_{3}=C_{4} .$ Find the condition for balance.
The Wien bridge is used in tuned amplifiers and in oscillators as well as for measuring or monitoring a frequency. To measure a frequency, one changes $R_{3}$ and $R_{4}$ simultaneously until $R_{3} \omega C_{3}$ is equal to unity.

Abhishek Jana

Numerade Educator

01:25

Problem 13

It is often necessary to shift the phase of a signal. Figure $7-28$ shows a simple circuit for doing this without affecting the amplitude of the signal. The resistances are adjustable, but equal. Use the polarities shown. They mean that $V_{i}$ is the voltage of the top terminal with respect to the bottom one, and $V_{0}$, is the voltage of the right-hand terminal with respect to the left-hand one.
(a) Show that $V_{o} / V_{i}=\exp \{2 j \arctan [1 /(R \omega C)]\}$.
(b) Draw a graph of the phase of $V_{o}$ with respect to $V_{i}$ in the range $R \omega C=0.1$ to 10 . Use a logarithmic scale for $R \omega C$.

James Kiss

Numerade Educator

02:57

Problem 14

From many points of view, alternating current is much preferable to direct current for power distribution. However, line losses are lower with direct current.
On a high-voltage overhead transmission line, the maximum operating voltage depends on several factors, such as corona losses (current losses through ionization of the air), the size of the insulators, etc. So, for a given line, the instantaneous voltage between one conductor and ground must never exceed a certain value, say $V_{0}$. Otherwise, the downtime and the cost of maintenance become excessive. The cost of a line increases rapidly with its voltage rating.

The current in the line can be made nearly as large as one likes without damaging it, since the conductors are well cooled by the ambient air. However, the power loss increases as the square of the current. So, the lower the current, the better.

With direct current, two conductors operate at $+V_{0}$ and $-V_{0}$ with respect to ground. The power delivered to the load is $2 V_{0} I_{\mathrm{dc}}$. With single-phase (SP) alternating current, there are two wires at $+V_{0} \cos \omega t$ and $-V_{0} \cos \omega t$.
(a) Show that for the same power at the load, the rms current $I_{\mathrm{SP}}$ is $2^{1 / 2} I_{\mathrm{dc}}$. The $I^{2} R$ losses in the line with single-phase alternating current are twice as large as those with direct current.
(b) With three-phase (TP) alternating current, we have three wires at $V_{0} \cos \omega t, V_{0} \cos (\omega t+2 \pi / 3)$, and $V_{0} \cos (\omega t+4 \pi / 3)$. We assume that the three load resistances connected between these wires and ground are equal. Then the current in the ground wire is zero. Show that for the same total power delivered to the three load resistances, the rms currents $I_{\mathrm{TP}}$ are $\left(\frac{2}{3}\right) 2^{1 / 2} I_{\mathrm{dc}} \approx I_{\mathrm{dc}}$.

With three-phase alternating current, the rms currents are thus about the same as with direct current, but there are three current-carrying wires instead of two, so that the losses are $50 \%$ larger than with direct current. A three-wire line is also more expensive than a two-wire one.

Salamat Ali

Numerade Educator