Problem Description:
The given circuit consists of a 9 V voltage source and three resistors (\( R_1 = 10 \, \text{k}\Omega \), \( R_2 = 2 \, \text{k}\Omega \), and \( R_3 = 1 \, \text{k}\Omega \)) connected in parallel. The task is to determine the total current (\( I_{\text{total}} \)) flowing through the circuit and the individual currents (\( I_{R1} \), \( I_{R2} \), and \( I_{R3} \)) through each resistor.
Solution:
#### Step 1: Understand the Circuit Configuration
The resistors \( R_1 \), \( R_2 \), and \( R_3 \) are connected in parallel across the voltage source. In a parallel circuit:
- The voltage across each resistor is the same and equal to the supply voltage (\( V = 9 \, \text{V} \)).
- The total current is the sum of the currents through each resistor.
#### Step 2: Calculate the Current Through Each Resistor
The current through each resistor can be calculated using Ohm's Law:
\[
I = \frac{V}{R}
\]
where:
- \( V \) is the voltage across the resistor,
- \( R \) is the resistance of the resistor.
##### For \( R_1 \):
\[
I_{R1} = \frac{V}{R_1} = \frac{9 \, \text{V}}{10 \, \text{k}\Omega} = \frac{9}{10 \times 10^3} = 0.9 \, \text{mA}
\]
##### For \( R_2 \):
\[
I_{R2} = \frac{V}{R_2} = \frac{9 \, \text{V}}{2 \, \text{k}\Omega} = \frac{9}{2 \times 10^3} = 4.5 \, \text{mA}
\]
##### For \( R_3 \):
\[
I_{R3} = \frac{V}{R_3} = \frac{9 \, \text{V}}{1 \, \text{k}\Omega} = \frac{9}{1 \times 10^3} = 9 \, \text{mA}
\]
#### Step 3: Calculate the Total Current
The total current \( I_{\text{total}} \) is the sum of the currents through all the resistors:
\[
I_{\text{total}} = I_{R1} + I_{R2} + I_{R3}
\]
Substitute the values:
\[
I_{\text{total}} = 0.9 \, \text{mA} + 4.5 \, \text{mA} + 9 \, \text{mA} = 14.4 \, \text{mA}
\]
#### Step 4: Verify Using the Equivalent Resistance
The equivalent resistance (\( R_{\text{eq}} \)) of resistors in parallel is given by:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]
Substitute the values:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{10 \, \text{k}\Omega} + \frac{1}{2 \, \text{k}\Omega} + \frac{1}{1 \, \text{k}\Omega}
\]
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{10 \times 10^3} + \frac{1}{2 \times 10^3} + \frac{1}{1 \times 10^3}
\]
\[
\frac{1}{R_{\text{eq}}} = 0.1 \times 10^{-3} + 0.5 \times 10^{-3} + 1 \times 10^{-3}
\]
\[
\frac{1}{R_{\text{eq}}} = 1.6 \times 10^{-3}
\]
\[
R_{\text{eq}} = \frac{1}{1.6 \times 10^{-3}} = 625 \, \Omega
\]
Now, calculate the total current using the equivalent resistance:
\[
I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{9 \, \text{V}}{625 \, \Omega} = 14.4 \, \text{mA}
\]
This confirms our previous calculation.
Final Answer:
\[
\boxed{I_{\text{total}} = 14.4 \, \text{mA}, \, I_{R1} = 0.9 \, \text{mA}, \, I_{R2} = 4.5 \, \text{mA}, \, I_{R3} = 9 \, \text{mA}}
\]
Parent Tip: Review the logic above to help your child master the concept of parallel circuit diagram.
