Certainly! Let's complete the table with the details provided.
We have three bulbs with the following characteristics:
### Bulbs' Individual Characteristics:
- Bulb 1:
- Voltage ([tex]\(V\)[/tex]): 3 volts
- Current ([tex]\(I\)[/tex]): 0.45 amperes
- Resistance ([tex]\(R\)[/tex]): 6.6666667 ohms
- Bulb 2:
- Voltage ([tex]\(V\)[/tex]): 3 volts
- Current ([tex]\(I\)[/tex]): 0.45 amperes
- Resistance ([tex]\(R\)[/tex]): 6.6666667 ohms
- Bulb 3:
- Voltage ([tex]\(V\)[/tex]): -3 volts
- Current ([tex]\(I\)[/tex]): 0.45 amperes
- Resistance ([tex]\(R\)[/tex]): -6.666667 ohms
### Total Characteristics:
1. Voltage ([tex]\(V_{\text{total}}\)[/tex]):
- Assuming the battery voltage or electric potential difference remains constant, the total voltage is 3 volts.
2. Current ([tex]\(I_{\text{total}}\)[/tex]):
- The total current remains as 0.45 amperes, which might suggest that the current through each bulb remains unchanged overall.
3. Resistance ([tex]\(R_{\text{total}}\)[/tex]):
- The total resistance ([tex]\(R_t\)[/tex]) can be calculated by summing the resistances of the individual bulbs:
[tex]\[
R_t = R_1 + R_2 + R_3 = 6.6666667 + 6.6666667 + (-6.666667) = 6.6666664 \text{ ohms}
\][/tex]
Putting this information into the table:
\begin{tabular}{|l|l|l|l|}
\hline & \multicolumn{1}{|c|}{ Voltage (v) } & \multicolumn{1}{|c|}{ Current (i) } & \multicolumn{1}{|c|}{ Resistance (R) } \\
\hline Bulb 1 & 3 & 0.45 & 6.6666667 \\
\hline Bulb 2 & 3 & 0.45 & 6.6666667 \\
\hline Bulb 3 & -3 & 0.45 & -6.666667 \\
\hline Total & 3 & 0.45 & 6.6666664 \\
\hline
\end{tabular}
To finalize, we have filled the 'Total' row as follows, based on the provided characteristics and calculations:
- Total Voltage (V): 3 volts
- Total Current (I): 0.45 amperes
- Total Resistance (R): 6.6666664 ohms
This completes the table with a detailed, step-by-step solution.
