Question

Chapter 18, Problem 1

A 47- resistor and a 29- resistor are connected in series across a 24-V battery. What is the voltage across (a) the 47- resistor and (b) the 29- resistor?

Chapter 18, Problem 2

The current in a 63- resistor is 0.077 A. This resistor is in series with a 42- resistor, and the series combination is connected across a battery. What is the battery voltage?

Chapter 18, Problem 3

What resistance must be placed in parallel with a 172- resistor to make the equivalent resistance 111 ?

Chapter 18, Problem 4

Two resistors, 59 and 82 , are connected in parallel. The current through the 82- resistor is 2.8 A. (a) Determine the current in the other resistor. (b) What is the total power consumed by the two resistors?

Chapter 18, Problem 5

Find the equivalent resistance between points A and B in the drawing.

Chapter 18, Problem 6

Find the equivalent resistance between points A and B in the drawing.

Chapter 18, Problem 7

A 68- resistor is connected in parallel with a 151- resistor. This parallel group is connected in series with a 25- resistor. The total combination is connected across a 23-V battery. Find (a) the current and (b) the power dissipated in the 151- resistor.

Chapter 18, Problem 8

A new "D" battery has an emf of 1.5 V. When a wire of negligible resistance is connected between the terminals of the battery, a current of 24 A is produced. Find the internal resistance of the battery.

Chapter 18, Problem 9

Two batteries, each with an internal resistance of 0.015 , are connected as in the drawing. In effect, the 9.0-V battery is being used to charge the 8.0-V battery. What is the current in the circuit?

Chapter 18, Problem 10

Determine the magnitude of the current in the (a) 8.0- and (b) 2.0- resistors in the drawing.

Chapter 18, Problem 11

An electronic flash attachment for a camera produces a flash by using the energy stored in a 890-F capacitor. Between flashes, the capacitor recharges through a resistor whose resistance is chosen so the capacitor recharges with a time constant of 3.9 s. Determine the value of the resistance.

Chapter 18, Problem 12

The circuit in the drawing contains two resistors and two capacitors that are connected to a battery via a switch. When the switch is closed, the capacitors begin to charge up. What is the time constant for the charging process?

Answer 1

1)

(a)

i = V/(R1+R2) = 24/(47+29) = 0.316 A

across 47 = V1 = i*R1 = 0.316*47 = 14.852 v

(b) across 29 = V2 = i*R2 = 0.316*29 = 9.164 V

(2)

V = i*(R1+R2) = 0.077*(63+42) = 8.805 V

(3)

1/Req = 1/R1 + 1/R2

1/111 = 1/172 + 1/R2

R2 = 312.9

(4)

(a)

in series current is same

through 59 = 2.8 A

(b)

P = i^2(R1+R2) = 2.8^2*(59+82) = 1105.44 W

(7)

Req = (151*68)/(151+68) + 25 = 72 ohms

I = V/Req = 23/72 = 0.32 A

current through 151 = (I*68)/(151+68) = 0.099 A

power = 0.099^2*151 = 1.48 W <<---answer