Let's solve the given problem step by step to understand all the missing readings. We already have some values given, and we need to compute the missing ones.
1. Given Readings:
- [tex]\( A_1 = 0.5 \, \text{A} \)[/tex]
- [tex]\( A_3 = 2 \, \text{A} \)[/tex]
- [tex]\( V_1 = 3 \, \text{V} \)[/tex]
2. Calculate [tex]\( A_2 \)[/tex]:
- According to the problem, [tex]\( A_2 \)[/tex] can be found using the relationship among the currents.
- [tex]\( A_2 = A_3 - A_1 \)[/tex]
- Substituting the given values:
[tex]\[
A_2 = 2 \, \text{A} - 0.5 \, \text{A} = 1.5 \, \text{A}
\][/tex]
3. Calculate [tex]\( V_2 \)[/tex]:
- Assuming a voltage divider in series, the voltage is divided proportionally to the current values.
- [tex]\( V_2 = \left( \frac{A_2}{A_3} \right) \cdot V_1 \)[/tex]
- Substituting the known values:
[tex]\[
V_2 = \left( \frac{1.5 \, \text{A}}{2 \, \text{A}} \right) \cdot 3 \, \text{V} = 2.25 \, \text{V}
\][/tex]
4. Calculate [tex]\( V_3 \)[/tex]:
- Since [tex]\( V_1 \)[/tex] is the total voltage and [tex]\( V_2 \)[/tex] is the part of it, [tex]\( V_3 \)[/tex] is the remaining part of the voltage.
- [tex]\( V_3 = V_1 - V_2 \)[/tex]
- Substituting the known values:
[tex]\[
V_3 = 3 \, \text{V} - 2.25 \, \text{V} = 0.75 \, \text{V}
\][/tex]
Having completed the calculations, the missing readings [tex]\( A_2 \)[/tex], [tex]\( V_2 \)[/tex], and [tex]\( V_3 \)[/tex] are:
[tex]\[
\begin{array}{|c|c|l|}
\hline
\text{Reading} & \text{Value with unit} & \text{Reason: Number or phrase} \\
\hline
A_1 & 0.5 \, \text{A} & \text{Given.} \\
\hline
A_2 & 1.5 \, \text{A} & 2 \, \text{A} - 0.5 \, \text{A} = 1.5 \, \text{A} \\
\hline
A_3 & 2 \, \text{A} & \text{Given.} \\
\hline
V_1 & 3 \, \text{V} & \text{Voltage of cell.} \\
\hline
V_2 & 2.25 \, \text{V} & \left( \frac{1.5 \, \text{A}}{2 \, \text{A}} \right) \cdot 3 \, \text{V} = 2.25 \, \text{V} \\
\hline
V_3 & 0.75 \, \text{V} & 3 \, \text{V} - 2.25 \, \text{V} = 0.75 \, \text{V} \\
\hline
\end{array}
\][/tex]
Therefore, the missing readings are:
- [tex]\( A_2 = 1.5 \, \text{A} \)[/tex]
- [tex]\( V_2 = 2.25 \, \text{V} \)[/tex]
- [tex]\( V_3 = 0.75 \, \text{V} \)[/tex]
This completes the solution.
