Given the network shown in Fig. 2.31: (a) find the required value for the resistor R; (b) use Table 2.1 to select a standard 10% tolerance resistor for R; (c) using the resistor selected in (b), determine the voltage across the 3.9-kΩ resistor; (d) calculate the percent error in the voltage V1,

Question

Given the network shown in Fig. 2.31: (a) find the required value for the resistor R; (b) use Table 2.1 to select a standard 10% tolerance resistor for R; (c) using the resistor selected in (b), determine the voltage across the 3.9-kΩ resistor; (d) calculate the percent error in the voltage [latex]V_{1},[/latex] if the standard resistor selected in (b) is used; and (e) determine the power rating for this standard component.

Accepted Answer

a. Using KVL, the voltage across R is 19 V. Then using Ohm’s law, the current in the loop is

I = 5/3.9k = 1.282 mA

The required value of R is then

R = 19/0.001282 = 14.82 kΩ

b. As shown in Table 2.1, the nearest standard 10% tolerance resistor is 15 kΩ.

TABLE 2.1 Standard resistor values for 5% and 10% tolerances (values available with a 10% tolerance shown in boldface)

1.0	10	100	1.0k	10k	100k	1.0M	10M
1.1	11	110	1.1k	11k	110k	1.1M	11M
1.2	12	120	1.2k	12k	120k	1.2M	12M
1.3	13	130	1.3k	13k	130k	1.3M	13M
1.5	15	150	1.5k	15k	150k	1.5M	15M
1.6	16	160	1.6k	16k	160k	1.6M	16M
1.8	18	180	1.8k	18k	180k	1.8M	18M
2.0	20	200	2.0k	20k	200k	2.0M	20M
2.2	22	220	2.2k	22k	220k	2.2M	22M
2.4	24	240	2.4k	24k	240k	2.4M
2.7	27	270	2.7k	27k	270k	2.7M
3.0	30	300	3.0k	30k	300k	3.0M
3.3	33	330	3.3k	33k	330k	3.3M
3.6	36	360	3.6k	36k	360k	3.6M
3.9	39	390	3.9k	39k	390k	3.9M
4.3	43	430	4.3k	43k	430k	4.3M
4.7	47	470	4.7k	47k	470k	4.7M
5.1	51	510	5.1k	51k	510k	5.1M
5.6	56	560	5.6k	56k	560k	5.6M
6.2	62	620	6.2k	62k	620k	6.2M
6.8	68	680	6.8k	68k	680k	6.8M
7.5	75	750	7.5k	75k	750k	7.5M
8.2	82	820	8.2k	82k	820k	8.2M
9.1	91	910	9.1k	91k	910k	9.1M

c. Using the standard 15-kΩ resistor, the actual current in the circuit is

I = 24/18.9k = 1.2698 mA

and the voltage across the 3.9-kΩ resistor is

V = IR = (0.0012698)(3.9k) = 4.952 V

d. The percent error involved in using the standard resistor is

% Error = (4.952 - 5)/5 × 100 = –0.96%

e. The power absorbed by the resistor R is then

P = IR = [latex](0.0012698)^{2}[/latex](15k) = 24.2 mW

Therefore, even a quarter-watt resistor is adequate in this application.

Question 2.23: Given the network shown in Fig. 2.31: (a) find the required ...

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