Sure, let's complete the table step-by-step.
1. Bulb 1: The voltage across Bulb 1 is [tex]\( V_1 = 10 \)[/tex] volts, the resistance is [tex]\( R_1 = 5 \)[/tex] ohms, and the current is [tex]\( i_1 = 2.0 \)[/tex] amperes.
2. Bulb 2: The voltage across Bulb 2 is [tex]\( V_2 = 8 \)[/tex] volts, the resistance is [tex]\( R_2 = 4 \)[/tex] ohms, and the current is [tex]\( i_2 = 2.0 \)[/tex] amperes.
3. Bulb 3: The voltage across Bulb 3 is [tex]\( V_3 = 12 \)[/tex] volts, the resistance is [tex]\( R_3 = 6 \)[/tex] ohms, and the current is [tex]\( i_3 = 2.0 \)[/tex] amperes.
To find the total voltage, sum the voltages of each bulb:
[tex]\[ V_{\text{total}} = V_1 + V_2 + V_3 = 10 + 8 + 12 = 30 \text{ volts} \][/tex]
Next, let's find the total resistance:
Using the formula for parallel resistance:
[tex]\[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{5} + \frac{1}{4} + \frac{1}{6} \][/tex]
[tex]\[ \frac{1}{R_{\text{total}}} \approx 0.6216216216216217 \][/tex]
Thus,
[tex]\[ R_{\text{total}} \approx 1.6216216216216215 \text{ ohms} \][/tex]
Finally, the total current using Ohm's law:
[tex]\[ i_{\text{total}} = \frac{V_{\text{total}}}{R_{\text{total}}} \][/tex]
[tex]\[ i_{\text{total}} = \frac{30}{1.6216216216216215} \approx 18.5 \text{ amperes} \][/tex]
Now, we can fill in the table:
[tex]\[
\begin{tabular}{|l|l|l|l|}
\hline & Voltage (V) & Current (i) & Resistance (R) \\
\hline Bulb 1 & 10 & 2.0 & 5 \\
\hline Bulb 2 & 8 & 2.0 & 4 \\
\hline Bulb 3 & 12 & 2.0 & 6 \\
\hline Total & 30 & 18.5 & 1.6216216216216215 \\
\hline
\end{tabular}
\][/tex]
