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The current in the following circuit was measured and found to be 2 mA. R1 R2 2k Ohm 3k Ohm .12V R3

The current in the following circuit was measured and found to be 2 mA. R1 R2 2k Ohm 3k Ohm .12V R3

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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### Circuit Analysis Example The current in the following circuit was measured and found to be **2 mA**. ![Circuit Diagram](image) This circuit diagram includes: - A 12V battery. - Three resistors connected in series, labeled as: - R1 = 2 kΩ - R2 = 3 kΩ - R3 = *Not specifically labeled for value in the text*. **a. How much power is being delivered by the battery? (Please provide the answer in Engineering Notation.)** To find the power being delivered by the battery, we first use Ohm's Law to calculate the total resistance in the circuit: 1. **Calculate Total Resistance (R_total):** Since the resistors are in series, the total resistance is the sum of all individual resistances. \[ R_{total} = R1 + R2 + R3 \] Assume \( R3 \) is known from the context or previous problems. For example, if \( R3 \) is approximately 1 kΩ, you get: \[ R_{total} = 2\,k\Omega + 3\,k\Omega + 1\,k\Omega = 6\,k\Omega \] 2. **Calculate Power (P):** Using the formula for power, \[ P = V \times I \] where \( V \) is the voltage of the battery (12V) and \( I \) is the current (2 mA or 0.002 A): \[ P = 12\,V \times 0.002\,A = 0.024\,W = 24\,mW \] Therefore, the power being delivered by the battery is **24 mW**. (Note: If \( R3 \) has a different value you can substitute the correct value in the equations above to find the appropriate total resistance and power delivered.)
Transcribed Image Text:### Circuit Analysis Example The current in the following circuit was measured and found to be **2 mA**. ![Circuit Diagram](image) This circuit diagram includes: - A 12V battery. - Three resistors connected in series, labeled as: - R1 = 2 kΩ - R2 = 3 kΩ - R3 = *Not specifically labeled for value in the text*. **a. How much power is being delivered by the battery? (Please provide the answer in Engineering Notation.)** To find the power being delivered by the battery, we first use Ohm's Law to calculate the total resistance in the circuit: 1. **Calculate Total Resistance (R_total):** Since the resistors are in series, the total resistance is the sum of all individual resistances. \[ R_{total} = R1 + R2 + R3 \] Assume \( R3 \) is known from the context or previous problems. For example, if \( R3 \) is approximately 1 kΩ, you get: \[ R_{total} = 2\,k\Omega + 3\,k\Omega + 1\,k\Omega = 6\,k\Omega \] 2. **Calculate Power (P):** Using the formula for power, \[ P = V \times I \] where \( V \) is the voltage of the battery (12V) and \( I \) is the current (2 mA or 0.002 A): \[ P = 12\,V \times 0.002\,A = 0.024\,W = 24\,mW \] Therefore, the power being delivered by the battery is **24 mW**. (Note: If \( R3 \) has a different value you can substitute the correct value in the equations above to find the appropriate total resistance and power delivered.)
b. What is the voltage drop across R1? c. What is the voltage drop across R2? d. What is the resistance value of R3?
Transcribed Image Text:b. What is the voltage drop across R1? c. What is the voltage drop across R2? d. What is the resistance value of R3?
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