In the circuit shown, the emf of the battery is 12.0 V, and each resistor has a resistance of 200.0 Q. Find the current supplied by the battery to this circuit. Answer Units: [A] R R www F & R
In the circuit shown, the emf of the battery is 12.0 V, and each resistor has a resistance of 200.0 Q. Find the current supplied by the battery to this circuit. Answer Units: [A] R R www F & R
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Example Problem: Calculating Current in a Resistor Circuit**
In the circuit shown, the emf of the battery is 12.0 V, and each resistor has a resistance of 200.0 Ω.
**Find the current supplied by the battery to this circuit.**
**Answer Units: [A]**
**Diagram Explanation:**
The provided diagram shows a circuit with a battery (denoted by ε) supplying a voltage of 12.0 V and several resistors. The resistors, marked as R and each with a resistance of 200.0 Ω, are arranged in a specific configuration. They form two parallel branches each containing two resistors in series. The current (I) supplied by the battery flows through this entire setup.
To solve for the current supplied by the battery (I), we need to analyze the resistances in the circuit:
1. **First step:** Calculate the equivalent resistance of one branch (two resistors in series):
\[ R_{\text{series}} = 200.0 \: \Omega + 200.0 \: \Omega = 400.0 \: \Omega \]
2. **Second step:** Since there are two of these series branches in parallel, calculate the equivalent resistance of the parallel combination:
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_{\text{series}}} + \frac{1}{R_{\text{series}}} \]
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{400.0 \: \Omega} + \frac{1}{400.0 \: \Omega} \]
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{400.0 \: \Omega} + \frac{1}{400.0 \: \Omega} = \frac{2}{400.0 \: \Omega} = \frac{1}{200.0 \: \Omega} \]
\[ R_{\text{parallel}} = 200.0 \: \Omega \]
3. **Third step:** Use Ohm's Law \( V = IR \) to find the current supplied by the battery:
\[ I = \frac{V}{R_{\text{parallel}}} = \frac{12.0 \: V}{200.0 \: \Omega} = 0.06 \: A \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F203b58df-0ca9-47a6-a660-754b46a1e0bb%2F3a2213c2-567e-41a2-b8b1-48fd3f532fe0%2Fprlgkx8_processed.png&w=3840&q=75)
Transcribed Image Text:**Example Problem: Calculating Current in a Resistor Circuit**
In the circuit shown, the emf of the battery is 12.0 V, and each resistor has a resistance of 200.0 Ω.
**Find the current supplied by the battery to this circuit.**
**Answer Units: [A]**
**Diagram Explanation:**
The provided diagram shows a circuit with a battery (denoted by ε) supplying a voltage of 12.0 V and several resistors. The resistors, marked as R and each with a resistance of 200.0 Ω, are arranged in a specific configuration. They form two parallel branches each containing two resistors in series. The current (I) supplied by the battery flows through this entire setup.
To solve for the current supplied by the battery (I), we need to analyze the resistances in the circuit:
1. **First step:** Calculate the equivalent resistance of one branch (two resistors in series):
\[ R_{\text{series}} = 200.0 \: \Omega + 200.0 \: \Omega = 400.0 \: \Omega \]
2. **Second step:** Since there are two of these series branches in parallel, calculate the equivalent resistance of the parallel combination:
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_{\text{series}}} + \frac{1}{R_{\text{series}}} \]
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{400.0 \: \Omega} + \frac{1}{400.0 \: \Omega} \]
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{400.0 \: \Omega} + \frac{1}{400.0 \: \Omega} = \frac{2}{400.0 \: \Omega} = \frac{1}{200.0 \: \Omega} \]
\[ R_{\text{parallel}} = 200.0 \: \Omega \]
3. **Third step:** Use Ohm's Law \( V = IR \) to find the current supplied by the battery:
\[ I = \frac{V}{R_{\text{parallel}}} = \frac{12.0 \: V}{200.0 \: \Omega} = 0.06 \: A \]
![**Question from Circuit Analysis Section**
---
**Continuing from the previous question:**
**Task:** Find the current through the lower two resistors.
**Answer Units:** [A]
---
Explanation:
In this step, you are required to determine the electric current flowing through the two resistors that are positioned at the lower part of the circuit diagram provided in the previous question. Ensure to use the relevant equations and principles such as Ohm's Law and the rules for series and parallel circuits. Your answer should be formatted in amperes [A].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F203b58df-0ca9-47a6-a660-754b46a1e0bb%2F3a2213c2-567e-41a2-b8b1-48fd3f532fe0%2Fi0wy1ji_processed.png&w=3840&q=75)
Transcribed Image Text:**Question from Circuit Analysis Section**
---
**Continuing from the previous question:**
**Task:** Find the current through the lower two resistors.
**Answer Units:** [A]
---
Explanation:
In this step, you are required to determine the electric current flowing through the two resistors that are positioned at the lower part of the circuit diagram provided in the previous question. Ensure to use the relevant equations and principles such as Ohm's Law and the rules for series and parallel circuits. Your answer should be formatted in amperes [A].
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