For the circuit shown in the figure, what is the current through the 10- resistor? 16V ©2010 Parson Education, C 1 A 0.6 A 5 Ω ww ww 10 Q2 ww 15 Ω ww 6Ω
For the circuit shown in the figure, what is the current through the 10- resistor? 16V ©2010 Parson Education, C 1 A 0.6 A 5 Ω ww ww 10 Q2 ww 15 Ω ww 6Ω
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Transcribed Image Text:**Transcription for Educational Website**
For the circuit shown in the figure, what is the current through the 10-Ω resistor?
**Circuit Diagram:**
- **Voltage Source:** 16V
- **Resistors:**
- 5 Ω
- 10 Ω
- 15 Ω
- 6 Ω
**Diagram Description:**
- The circuit is powered by a 16V battery.
- A 5 Ω resistor is connected in series with a parallel combination of three resistors: 10 Ω, 15 Ω, and 6 Ω.
**Question:**
What is the current flowing through the 10 Ω resistor?
**Options:**
- 1 A
- 0.6 A
Please note: To solve the problem, you will need to calculate the equivalent resistance and apply Ohm's Law to determine the current through the 10 Ω resistor in the parallel section.
![### Circuit Diagram Analysis
This diagram represents a simple electrical circuit with the following components:
1. **Voltage Source**: A 16V battery is depicted on the left side of the diagram, providing electrical energy to the circuit.
2. **Resistors in Series**:
- A 10Ω resistor.
- A 15Ω resistor.
- A 6Ω resistor.
These resistors are connected in series along the circuit.
### Question
What is the total current flowing through the circuit?
#### Answer Options
- ○ 1 A
- ○ 0.6 A
- ○ 0.4 A
- ○ 1.6 A
- ○ 1.2 A
- ○ 2 A
### Explanation
To find the total current flowing through this series circuit, use Ohm’s Law, which states:
\[ I = \frac{V}{R} \]
Where:
- \( I \) is the current in amperes (A),
- \( V \) is the voltage in volts (V),
- \( R \) is the total resistance in ohms (Ω).
### Step-by-Step Calculation
1. **Calculate Total Resistance** (\( R \)):
- Since the resistors are in series, the total resistance \( R \) is the sum of each resistor.
- \( R = 10Ω + 15Ω + 6Ω = 31Ω \)
2. **Apply Ohm’s Law**:
- With a voltage \( V \) of 16V and a total resistance \( R \) of 31Ω:
- \( I = \frac{16V}{31Ω} \approx 0.516 A \)
Therefore, the closest answer to the calculated current is approximately \( 0.5 A \), which means none of the listed options precisely match. Adjust the question options accordingly based on precise calculations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8daeabfb-08fc-4e08-b5f3-009f3913f10d%2Fb21a602d-dec2-4ad6-8e7a-572ca656106e%2Fj8js77j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Circuit Diagram Analysis
This diagram represents a simple electrical circuit with the following components:
1. **Voltage Source**: A 16V battery is depicted on the left side of the diagram, providing electrical energy to the circuit.
2. **Resistors in Series**:
- A 10Ω resistor.
- A 15Ω resistor.
- A 6Ω resistor.
These resistors are connected in series along the circuit.
### Question
What is the total current flowing through the circuit?
#### Answer Options
- ○ 1 A
- ○ 0.6 A
- ○ 0.4 A
- ○ 1.6 A
- ○ 1.2 A
- ○ 2 A
### Explanation
To find the total current flowing through this series circuit, use Ohm’s Law, which states:
\[ I = \frac{V}{R} \]
Where:
- \( I \) is the current in amperes (A),
- \( V \) is the voltage in volts (V),
- \( R \) is the total resistance in ohms (Ω).
### Step-by-Step Calculation
1. **Calculate Total Resistance** (\( R \)):
- Since the resistors are in series, the total resistance \( R \) is the sum of each resistor.
- \( R = 10Ω + 15Ω + 6Ω = 31Ω \)
2. **Apply Ohm’s Law**:
- With a voltage \( V \) of 16V and a total resistance \( R \) of 31Ω:
- \( I = \frac{16V}{31Ω} \approx 0.516 A \)
Therefore, the closest answer to the calculated current is approximately \( 0.5 A \), which means none of the listed options precisely match. Adjust the question options accordingly based on precise calculations.
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