Sample Solution from
Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf
9th Edition
ISBN: 9781259989452
Chapter 2
Problem 1E
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Textbook Problem

Convert the following to engineering notation:

  1. (a) 0.045 W
  2. (b) 2000 pJ
  3. (c) 0.1 ns
  4. (d) 39,212 as
  5. (e) 3 Ω
  6. (f) 18,000 m
  7. (g) 2,500,000,000,000 bits
  8. (h) 1015 atoms/cm3

Expert Solution

(a)

To determine

Convert 0.045 W into engineering notation.

Answer

The conversion of 0.045 W into engineering notation is 45 mW.

Explanation of Solution

Given data:

The power given is 0.045 W.

Calculation:

The conversion of 0.045 W in engineering notation is as follows.

0.045 W=0.045×103 mW                 {∵1 W=103 mW}=45 mW

Conclusion:

Thus, the conversion of 0.045 W into engineering notation is 45 mW.

Expert Solution

(b)

To determine

Convert 2000 pJ into engineering notation.

Answer

The conversion of 2000 pJ into engineering notation is 2 nJ.

Explanation of Solution

Given data:

The energy given is 2000 pJ.

Calculation:

The conversion of 2000 pJ in engineering notation is as follows.

2000 pJ=2000×10−3 nJ                 {∵1 pJ=10−3 nJ}=2 nJ

Conclusion:

Thus, the conversion of 2000 pJ into engineering notation is 2 nJ.

Expert Solution

(c)

To determine

Convert 0.1 ns into engineering notation.

Answer

The conversion of 0.1 ns into engineering notation is 100 ps.

Explanation of Solution

Given data:

The time given is 0.1 ns.

Calculation:

The conversion of 0.1 ns in engineering notation is as follows.

0.1 ns=0.1×103 ps                 {∵1 ns=103 ps}=100 ps

Conclusion:

Thus, the conversion of 0.1 ns into engineering notation is 100 ps.

Expert Solution

(d)

To determine

Convert 39212 as into engineering notation.

Answer

The conversion of 39212 as into engineering notation is 39.212 fs.

Explanation of Solution

Given data:

The time given is 39212 as.

Calculation:

The conversion of 39212 as in engineering notation is as follows.

39212 as=39212×10−3 fs                 {∵1 as=10−3 fs}=39.212 fs

Conclusion:

Thus, the conversion of 39212 as into engineering notation is 39.212 fs.

Expert Solution

(e)

To determine

Convert 3 Ω into engineering notation.

Answer

The conversion of 3 Ω into engineering notation is 3 Ω.

Explanation of Solution

Given data:

The resistance given is 3 Ω.

The given data is already in engineering notation.

Conclusion:

Thus, the conversion of 3 Ω into engineering notation is 3 Ω.

Expert Solution

(f)

To determine

Convert 18000 m into engineering notation.

Answer

The conversion of 18000 m into engineering notation is 18 km.

Explanation of Solution

Given data:

The distance given is 18000 m.

Calculation:

The conversion of 18000 m in engineering notation is as follows.

18000 m=18000×10−3 km                 {∵1 m=10−3 km}=18 km

Conclusion:

Thus, the conversion of 18000 m into engineering notation is 18 km.

Expert Solution

(g)

To determine

Convert 2,500,000,000,000 bits into engineering notation.

Answer

The conversion of 2,500,000,000,000 bits into engineering notation is 2.5 terabits.

Explanation of Solution

Given data:

The memory given is 2,500,000,000,000 bits.

Calculation:

The conversion of 2,500,000,000,000 bits in engineering notation is as follows.

2,500,000,000,000 bits=(2,500,000,000,000×10−12 terabits)                 {∵1 bit=10−12 terabits}=2.5 terabits

Conclusion:

Thus, the conversion of 2,500,000,000,000 bits into engineering notation is 2.5 terabits.

Expert Solution

(h)

To determine

Convert 1015 atomscm3 into engineering notation.

Answer

The conversion of 1015 atomscm3 into engineering notation is 1021 atoms m3.

Explanation of Solution

Given data:

The volume density of atom given is 1015 atomscm3.

Calculation:

The conversion of 1015 atomscm3 in engineering notation is as follows.

1015 atomscm3=1015 atoms10−6 m3                 {∵1 cm3=10−6 m3}=1021 atoms m3  

Conclusion:

Thus, the conversion of 1015 atomscm3 into engineering notation is 1021 atoms m3.

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