The circuit below applies to Problems 18-20. Hint: V=1(R1+ R2||R3) I 50 12 R1 R2R3 R2||R3 R2+R3 21V R2 30 60 R3 R2 -- 12= R2+R3 V = IR (Ohm's Law) 18. Find Ij. a. 1.0 A b. 1.3 А с. 1.7 A d. 2.0 A 19. Find Iz. a. 1.0 A b. 1.3 А с. 1.7 А d. 2.0 A 20. Find the power absorbed by the 60 resistor. Hint: P= (1;} (R;) а. 3 W b. 6 W c. 8 W d. 24 W
The circuit below applies to Problems 18-20. Hint: V=1(R1+ R2||R3) I 50 12 R1 R2R3 R2||R3 R2+R3 21V R2 30 60 R3 R2 -- 12= R2+R3 V = IR (Ohm's Law) 18. Find Ij. a. 1.0 A b. 1.3 А с. 1.7 A d. 2.0 A 19. Find Iz. a. 1.0 A b. 1.3 А с. 1.7 А d. 2.0 A 20. Find the power absorbed by the 60 resistor. Hint: P= (1;} (R;) а. 3 W b. 6 W c. 8 W d. 24 W
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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Transcribed Image Text:**Circuit Analysis Problem Set**
The following circuit applies to Problems 18-20. The circuit diagram consists of:
- A voltage source of 21V.
- Three resistors: R1 = 5Ω, R2 = 3Ω, R3 = 6Ω.
- A parallel combination of resistors R2 and R3.
**Relevant Equations:**
1. Parallel resistors:
\( R_2 || R_3 = \frac{R_2 R_3}{R_2 + R_3} \)
2. Current through R2 (\( I_2 \)):
\( I_2 = \frac{R_2}{R_2 + R_3} I \)
3. Ohm's Law:
\( V = IR \)
**Graph Explanation:**
- The circuit depicts a series-parallel configuration.
- Current \( I \) flows through R1, then splits into \( I_1 \) through R2 and \( I_2 \) through R3.
**Problems:**
18. **Find \( I_1 \):**
- a. 1.0 A
- b. 1.3 A
- c. 1.7 A
- d. 2.0 A
19. **Find \( I_2 \):**
- a. 1.0 A
- b. 1.3 A
- c. 1.7 A
- d. 2.0 A
20. **Find the power absorbed by the 6Ω resistor. Hint: \( P = (I_2)^2 (R_3) \):**
- a. 3 W
- b. 6 W
- c. 8 W
- d. 24 W
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