Chapter 1 Practice Problems

Practice Problems

Problem 1 — Current Calculation

A charge of 120 mC passes through a wire in 40 milliseconds.

(a) Calculate the current in amperes.

(b) Express the current in milliamperes.

(c) How long would it take for 1 coulomb of charge to pass at this rate?

Solution

(a) Convert units and apply the current definition:

\[I = \frac{Q}{t} = \frac{120 \times 10^{-3}\text{ C}}{40 \times 10^{-3}\text{ s}} = \frac{0.12}{0.04} = 3\text{ A}\]

(b) $3\text{ A} = 3000\text{ mA}$

(c) [t = \frac{Q}{I} = \frac{1\text{ C}}{3\text{ A}} = 0.333\text{ s} \approx 333\text{ ms}]

Problem 2 — Ohm's Law Application

A 9V battery is connected to a circuit containing a 3.3kΩ resistor.

(a) Calculate the current through the resistor.

(b) Calculate the power dissipated by the resistor.

(c) Is a standard 1/4W (0.25W) resistor adequate for this application?

Solution

(a) Apply Ohm's Law:

\[I = \frac{V}{R} = \frac{9\text{ V}}{3300\text{ Ω}} = 2.73\text{ mA}\]

(b) Calculate power:

\[P = VI = 9\text{ V} \times 0.00273\text{ A} = 24.5\text{ mW}\]

Or equivalently: $P = V^2/R = 81/3300 = 24.5\text{ mW}$

(c) 24.5 mW is well below the 250 mW rating of a 1/4W resistor. A 1/4W resistor is more than adequate — it is running at less than 10% of its rated power, which is excellent practice.

Problem 3 — Power and Energy

A 60W light bulb is left on for 8 hours.

(a) How much energy (in joules) is consumed?

(b) Express the energy in kilowatt-hours.

(c) If electricity costs $0.13/kWh, what does it cost to run this bulb for 8 hours?

(d) What current does the bulb draw from a 120V outlet?

Solution

(a) Convert 8 hours to seconds and calculate energy:

\[W = Pt = 60\text{ W} \times (8 \times 3600\text{ s}) = 60 \times 28800 = 1{,}728{,}000\text{ J} = 1.728\text{ MJ}\]

(b) $W = 60\text{ W} \times 8\text{ h} = 480\text{ Wh} = 0.48\text{ kWh}$

(c) Cost $= 0.48\text{ kWh} \times \$0.13/\text{kWh} = \$0.062 \approx 6.2\text{ cents}$

(d) Using $P = VI$:

\[I = \frac{P}{V} = \frac{60\text{ W}}{120\text{ V}} = 0.5\text{ A}\]

Problem 4 — Conductance

A resistor has a resistance of 470Ω.

(a) Calculate its conductance in siemens.

(b) If two 470Ω resistors are connected in parallel, what is the equivalent conductance?

(c) What is the equivalent resistance of the parallel combination?

Solution

(a) [G = \frac{1}{R} = \frac{1}{470\text{ Ω}} = 2.128\text{ mS}]

(b) Conductances in parallel add directly:

\[G_{total} = G_1 + G_2 = 2.128\text{ mS} + 2.128\text{ mS} = 4.255\text{ mS}\]

(c) [R_{eq} = \frac{1}{G_{total}} = \frac{1}{4.255 \times 10^{-3}} = 235\text{ Ω}]

This confirms the parallel resistance formula: two equal resistors in parallel give half the resistance of one.