Question

table 1: currents and voltages in circuit i
quantity | simulated value
i_t |
i_1 |
i_2 |
i_3 |
v_1 |
v_2 |
v_3 |
circuit has vs 12v, r1 6.8kω, r2 1kω, r3 3.3kω

Explanation:

Step1: Analyze the circuit structure

The circuit has a voltage source \( V_s = 12\,\text{V} \). Resistor \( R_1 = 6.8\,\text{k}\Omega \) is in parallel with the series combination of \( R_2 = 1\,\text{k}\Omega \) and \( R_3 = 3.3\,\text{k}\Omega \).

Step2: Calculate the equivalent resistance of \( R_2 \) and \( R_3 \) (series)

For series resistors, \( R_{23} = R_2 + R_3 \). Substituting values: \( R_{23} = 1\,\text{k}\Omega + 3.3\,\text{k}\Omega = 4.3\,\text{k}\Omega \).

Step3: Calculate the equivalent resistance of \( R_1 \) and \( R_{23} \) (parallel)

The formula for parallel resistors is \( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_{23}} \). Substituting \( R_1 = 6.8\,\text{k}\Omega \) and \( R_{23} = 4.3\,\text{k}\Omega \):
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{6.8\times10^3} + \frac{1}{4.3\times10^3} = \frac{4.3 + 6.8}{6.8\times4.3\times10^6} = \frac{11.1}{29.24\times10^6} \]
\[ R_{\text{eq}} = \frac{29.24\times10^6}{11.1\times10^3} \approx 2634.23\,\Omega \approx 2.63\,\text{k}\Omega \]

Step4: Calculate total current \( I_T \)

Using Ohm's Law \( I = \frac{V}{R} \), \( I_T = \frac{V_s}{R_{\text{eq}}} = \frac{12\,\text{V}}{2634.23\,\Omega} \approx 4.555\times10^{-3}\,\text{A} \approx 4.56\,\text{mA} \).

Step5: Calculate voltage across \( R_{23} \) (and thus across \( R_1 \), since they are in parallel)

\( V_{23} = V_1 = I_T \times R_{23} \)? Wait, no—wait, in parallel, \( V_1 = V_{23} = V_s \)? No, wait, no: the voltage source is across the parallel combination. Wait, actually, the voltage across \( R_1 \) and across \( R_{23} \) is equal to \( V_s = 12\,\text{V} \)? Wait, no, looking at the circuit diagram: the voltage source is connected across the parallel network of \( R_1 \) and ( \( R_2 \) series \( R_3 \) ). So yes, \( V_1 = V_{23} = V_s = 12\,\text{V} \)? Wait, that can't be, because the resistors are in parallel with the voltage source. Wait, no, the voltage source is the supply, so the voltage across \( R_1 \) is \( V_s = 12\,\text{V} \), and the voltage across \( R_2 \) and \( R_3 \) (series) is also \( 12\,\text{V} \).

Wait, I made a mistake earlier. Let's re-analyze:

- \( R_1 \) is in parallel with the series combination of \( R_2 \) and \( R_3 \). The voltage across \( R_1 \) ( \( V_1 \) ) and across \( R_2 + R_3 \) ( \( V_{23} \) ) is equal to the source voltage \( V_s = 12\,\text{V} \).

Step6: Calculate \( I_1 \) (current through \( R_1 \))

Using Ohm's Law: \( I_1 = \frac{V_1}{R_1} = \frac{12\,\text{V}}{6.8\times10^3\,\Omega} \approx 1.765\times10^{-3}\,\text{A} \approx 1.77\,\text{mA} \).

Step7: Calculate \( I_{23} \) (current through \( R_2 \) and \( R_3 \), which is \( I_2 = I_3 \) since they are in series)

\( I_{23} = \frac{V_{23}}{R_{23}} = \frac{12\,\text{V}}{4.3\times10^3\,\Omega} \approx 2.791\times10^{-3}\,\text{A} \approx 2.79\,\text{mA} \). So \( I_2 = I_3 = 2.79\,\text{mA} \).

Step8: Calculate total current \( I_T \)

\( I_T = I_1 + I_{23} \) (since they are in parallel). Substituting values: \( I_T = 1.77\,\text{mA} + 2.79\,\text{mA} = 4.56\,\text{mA} \), which matches the earlier calculation (considering rounding errors).

Step9: Calculate voltages \( V_2 \) and \( V_3 \)

For \( R_2 \) (series with \( R_3 \)): \( V_2 = I_2 \times R_2 = 2.791\times10^{-3}\,\text{A} \times 1\times10^3\,\Omega = 2.791\,\text{V} \approx 2.79\,\text{V} \).

For \( R_3 \): \( V_3 = I_3 \times R_3 = 2.791\times10^{-3}\,\text{A} \times 3.3\times10^3\,\Omega = 9.21\,\text{V} \) (check: \( V_2 + V_3 = 2.79 + 9.21 = 12\,\text{V} \), which matches \( V_{23} \)).

Now, summarizing:

- \( I_T \): Total current (sum of \( I_1 \) and \( I_{23} \)): \( \approx 4.56\,\text{mA} \)
- \( I_1 \): \( \approx 1.77\,\text{mA} \)
- \( I_2 = I_3 \): \( \approx 2.79\,\text{mA} \)
- \( V_1 \): \( 12\,\text{V} \) (since it's in parallel with the source)
- \( V_2 \): \( \approx 2.79\,\text{V} \)
- \( V_3 \): \( \approx 9.21\,\text{V} \)

Let's compute more precisely:

- \( I_1 = 12 / 6800 = 12000 / 6800 = 120 / 68 = 30 / 17 \approx 1.7647\,\text{mA} \)
- \( R_{23} = 1000 + 3300 = 4300\,\Omega \)
- \( I_{23} = 12 / 4300 \approx 2.7907\,\text{mA} \)
- \( I_T = I_1 + I_{23} = 30/17 + 12/4300 \). Let's compute with decimals: \( 1.7647 + 2.7907 = 4.5554\,\text{mA} \approx 4.56\,\text{mA} \)
- \( V_2 = I_2 \times R_2 = (12 / 4300) \times 1000 = 12000 / 4300 = 120 / 43 \approx 2.7907\,\text{V} \)
- \( V_3 = I_3 \times R_3 = (12 / 4300) \times 3300 = (12 \times 3300) / 4300 = (12 \times 33) / 43 = 396 / 43 \approx 9.2093\,\text{V} \)
- \( V_1 = 12\,\text{V} \) (exact, since it's across the source)

So filling the table:

| Quantity | Simulated Value (approximate) |
|----------|-------------------------------|
| \( I_T \) | \( \approx 4.56\,\text{mA} \) |
| \( I_1 \) | \( \approx 1.76\,\text{mA} \) |
| \( I_2 \) | \( \approx 2.79\,\text{mA} \) |
| \( I_3 \) | \( \approx 2.79\,\text{mA} \) |
| \( V_1 \) | \( 12\,\text{V} \) |
| \( V_2 \) | \( \approx 2.79\,\text{V} \) |
| \( V_3 \) | \( \approx 9.21\,\text{V} \) |

If we use more precise calculations:

- \( I_1 = 12 / 6800 = 0.0017647\,\text{A} = 1.7647\,\text{mA} \)
- \( I_{23} = 12 / 4300 = 0.0027907\,\text{A} = 2.7907\,\text{mA} \)
- \( I_T = 0.0017647 + 0.0027907 = 0.0045554\,\text{A} = 4.5554\,\text{mA} \)
- \( V_2 = 0.0027907 \times 1000 = 2.7907\,\text{V} \)
- \( V_3 = 0.0027907 \times 3300 = 9.2093\,\text{V} \)
- \( V_1 = 12\,\text{V} \) (exact)

Answer:

To fill the table, the values (rounded appropriately) are:

• \( I_T \): \( \approx 4.56\,\text{mA} \)
• \( I_1 \): \( \approx 1.76\,\text{mA} \)
• \( I_2 \): \( \approx 2.79\,\text{mA} \)
• \( I_3 \): \( \approx 2.79\,\text{mA} \)
• \( V_1 \): \( 12\,\text{V} \)
• \( V_2 \): \( \approx 2.79\,\text{V} \)
• \( V_3 \): \( \approx 9.21\,\text{V} \)

(For precise values, use the exact fractions or more decimal places as needed.)