2

Assessment a Problems

AP 2.1

[a] Note that the current ib is in the same circuit branch as the 8 A current source; however, ib is defined in the opposite direction of the current source. Therefore, ib = −8 A

Next, note that the dependent voltage source and the independent voltage source are in parallel with the same polarity. Therefore, their voltages are equal, and ib −8 vg =

=

= −2 V

4

4

[b] To find the power associated with the 8 A source, we need to find the voltage drop across the source, vi . Note that the two independent sources are in parallel, and that the voltages vg and v1 have the same polarities, so these voltages are equal: vi = vg = −2 V

Using the passive sign convention, ps = (8 A)(vi ) = (8 A)(−2 V) = −16 W

Thus the current source generated 16 W of power.

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval

2–1 system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,

Pearson Education, Inc., Upper Saddle River, NJ 07458.

2–2

CHAPTER 2. Circuit Elements

AP 2.2

[a] Note from the circuit that vx = −25 V. To find α note that the two current sources are in the same branch of the circuit but their currents flow in opposite directions. Therefore αvx = −15 A

Solve the above equation for α and substitute for vx, α= −15 A

−15 A

=

= 0.6 A/V vx −25 V

[b] To find the power associated with the voltage source we need to know the current, iv . Note that this current is in the same branch of the circuit as the dependent current source and these two currents flow in the same direction. Therefore, the current iv is the same as the current of the dependent source: iv = αvx = (0.6)(−25) = −15 A

Using the passive sign convention, ps = −(iv )(25 V) = −(−15 A)(25 V) = 375 W.

Thus the voltage source dissipates 375 W.

AP 2.3

[a] The resistor and the voltage source are in parallel and the resistor voltage and the voltage source have the same polarities. Therefore these two voltages are the same: vR = vg = 1 kV

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,

Pearson Education, Inc., Upper Saddle River, NJ 07458.

Problems

2–3

Note from the circuit that the current through the resistor is ig = 5 mA.

Use Ohm’s law to calculate the value of the resistor: vR 1 kV

R=

=

= 200 kΩ ig 5 mA

Using the passive sign convention to calculate the power in the resistor, pR = (vR )(ig ) = (1 kV)(5 mA) = 5 W

The resistor is dissipating 5 W of power.

[b] Note from part (a) the vR = vg and iR = ig . The power delivered by the source is thus psource −3 W psource = −vg ig so vg = −

=−

= 40 V ig 75 mA

Since we now have the value of both the voltage and the current for the resistor, we can use Ohm’s law to calculate the resistor value:

R=

40 V vg =

= 533.33 Ω ig 75 mA

The power absorbed by the resistor must equal the power generated by the source. Thus, pR = −psource = −(−3 W) = 3 W

[c] Again, note the iR = ig . The power dissipated by the resistor can be determined from the resistor’s current: pR = R(iR )2 = R(ig )2

Solving for ig , pr 480 mW

=

= 0.0016

R

300 Ω

Then, since vR = vg

i2g =

so

ig =

vR = RiR = Rig = (300 Ω)(40 mA) = 12 V

√

0.0016 = 0.04 A = 40 mA

so

vg = 12 V

AP 2.4

[a] Note from the circuit that the current through the conductance G is ig , flowing from top to bottom, because the current