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1. GIVEN THAT YOU HAVE A LOT OF STANDARD RESISTOR BUT WITH VALUE OF 10 k OHMS, 8 k OHMS, and 2 k OHMS ONLY. IF ALL VALUES MUST BE USED IN THE CIRCUIT, ASSIGN VALUES TO EACH RESISTOR ON THE CIRCUITS BELOW TO MEET THE REQUIRED TOTAL RESISTANCE OF 11 k OHMS. 2. SAME QUESTION ON NUMBER 1, BUT WITH REQUIRED TOTAL RESISTANCE OF 15 k OHMS. SEE ATTACHED PICTURE

1. GIVEN THAT YOU HAVE A LOT OF STANDARD RESISTOR BUT WITH VALUE OF 10 k OHMS, 8 k OHMS, and 2 k OHMS ONLY. IF ALL VALUES MUST BE USED IN THE CIRCUIT, ASSIGN VALUES TO EACH RESISTOR ON THE CIRCUITS BELOW TO MEET THE REQUIRED TOTAL RESISTANCE OF 11 k OHMS. 2. SAME QUESTION ON NUMBER 1, BUT WITH REQUIRED TOTAL RESISTANCE OF 15 k OHMS. SEE ATTACHED PICTURE

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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1. GIVEN THAT YOU HAVE A LOT OF STANDARD RESISTOR BUT WITH VALUE OF 10 k OHMS, 8 k OHMS, and 2 k OHMS ONLY. IF ALL VALUES MUST BE USED IN THE CIRCUIT, ASSIGN VALUES TO EACH RESISTOR ON THE CIRCUITS BELOW TO MEET THE REQUIRED TOTAL RESISTANCE OF 11 k OHMS. 2. SAME QUESTION ON NUMBER 1, BUT WITH REQUIRED TOTAL RESISTANCE OF 15 k OHMS. SEE ATTACHED PICTURE
GUEN that YOU HAUE OF IDK2, 8K2, AHD 2K2 OHLY IF ALL TESE VAWES nUST BE USED I CIRCUIT, ASI EN VALUES TO EACH RESIC TOR OH TE CIRCUITS BELOW TO MEET THE REQUIRED 1OTAU RECISTANCE. A LOT OF CTan taRD RESISTOR RT WITH VAWE KT-IKe RI RE= K4- 2) Some QeSTIDu LIKE GUESTIOH, vawES bx2, BK2, 2K52 RT = 15 K2 RI- 1 RA =
Transcribed Image Text:GUEN that YOU HAUE OF IDK2, 8K2, AHD 2K2 OHLY IF ALL TESE VAWES nUST BE USED I CIRCUIT, ASI EN VALUES TO EACH RESIC TOR OH TE CIRCUITS BELOW TO MEET THE REQUIRED 1OTAU RECISTANCE. A LOT OF CTan taRD RESISTOR RT WITH VAWE KT-IKe RI RE= K4- 2) Some QeSTIDu LIKE GUESTIOH, vawES bx2, BK2, 2K52 RT = 15 K2 RI- 1 RA =
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