**Educational Website Text**### Circuit Analysis Exercise**Objective**: Solve the circuit for the quantities listed.#### Circuit DiagramThe diagram is a simple series circuit consisting of:- A DC voltage source labeled $ E $.- Four resistors in series: $ R_1, R_2, R_3, $ and $ R_4 $.#### Given Values- $ E = 150 \, \text{VDC} $- $ R_1 = 2.7 \, \text{k}\Omega $- $ R_2 = 2.3 \, \text{k}\Omega $- $ R_3 = 6000 \, \Omega $- $ R_4 = 3000 \, \Omega $#### Required Calculations1. **Total Resistance ($ R_{\text{Total}} $)**2. **Total Current ($ I_{\text{Total}} $)**3. **Voltage across each resistor**: - $ V_{R_1} $ - $ V_{R_2} $ - $ V_{R_3} $ - $ V_{R_4} $4. **Total Power Consumption (Power Total)****Note**: You will need to use Ohm’s Law and the formulas for total resistance and power in a series circuit to complete these calculations.

Answered: Circuits for • Solve the folbwing…

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...

See similar textbooks

Related questions

Concept explainers

Question

**Educational Website Text**

### Circuit Analysis Exercise

**Objective**: Solve the circuit for the quantities listed.

#### Circuit Diagram
The diagram is a simple series circuit consisting of:

- A DC voltage source labeled $ E $.
- Four resistors in series: $ R_1, R_2, R_3, $ and $ R_4 $.

#### Given Values
- $ E = 150 \, \text{VDC} $
- $ R_1 = 2.7 \, \text{k}\Omega $
- $ R_2 = 2.3 \, \text{k}\Omega $
- $ R_3 = 6000 \, \Omega $
- $ R_4 = 3000 \, \Omega $

#### Required Calculations
1. **Total Resistance ($ R_{\text{Total}} $)**
2. **Total Current ($ I_{\text{Total}} $)**
3. **Voltage across each resistor**:
- $ V_{R_1} $
- $ V_{R_2} $
- $ V_{R_3} $
- $ V_{R_4} $
4. **Total Power Consumption (Power Total)**

**Note**: You will need to use Ohm’s Law and the formulas for total resistance and power in a series circuit to complete these calculations.

Expert Solution

Step 1

Solution:

Given,

$\begin{array}{rcl} R_{1} & = & 2.7 k Ω = 2700 Ω \\ R_{2} & = & 3.3 K Ω = 3300 Ω \\ R_{3} & = & 6000 Ω \\ R_{4} & = & 3000 Ω \\ a n d, E & = & 150 V \end{array}$

from the circuit,

(a)

$R_{t o t a l} = R_{1} + R_{2} + R_{3} + R_{4} (\sin c e a l l r e s i s t o r s i s i n s e r i e s) R_{t o t a l} = 2700 + 3300 + 6000 + 3000 R_{t o t a l} = 15000 Ω$

(b)

$I_{t o t a l} = \frac{E}{R_{t o t a l}} = \frac{150}{2700 + 3300 + 6000 + 3000} I_{t o t a l} = \frac{150}{15000} I_{t o t a l} = 0.01 A I_{t o t a l} = 10 m A$

(c)

$V_{R 1} = I_{t o t a l} \times R_{1} V_{R 1} = 0.01 \times 2700 V_{R 1} = 27 V$

(d)

$V_{R 2} = I_{t o t a l} \times R_{2} V_{R 2} = 0.01 \times 3300 V_{R 2} = 33 V$

(e)

$V_{R 3} = I_{t o t a l} \times R_{3} V_{R 3} = 0.01 \times 6000 V_{R 3} = 60 V$

(f)

$V_{R 4} = I_{t o t a l} \times R_{4} V_{R 4} = 0.01 \times 3000 V_{R 1} = 30 V$

(g)

$T o t a l p o w e r = I^{2} (R_{1} + R_{2} + R_{3} + R_{4}) T o t a l p o w e r = I^{2} \times R_{t o t a l} T o t a l p o w e r = 0 . 01^{2} \times 15000 T o t a l p o w e r = (1 \times 10^{- 4}) \times 15000 T o t a l p o w e r = 1.5 W a t t$

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.

Similar questions