4. Fill out the table for the circuit diagramed at the right. }(c) Circuit Position & Voltage (V) & Current (A) & Resistance (Omega) 1 & & & 2.00 2 & & & 5.00 3 & & & 10.0 Total & 9.00 & &

#Explanation#<br /><br />The table provided corresponds to a circuit with three resistors in series. The total voltage across the circuit is given as 9V. In a series circuit, the total resistance is the sum of the individual resistances, and the current through each component is the same.<br /><br />The total resistance ($R_{total}$) can be calculated using the formula:<br /><br />\[R_{total} = R_1 + R_2 + R_3\]<br /><br />Substituting the given values:<br /><br />\[R_{total} = 2\Omega + 5\Omega + 10\Omega = 17\Omega\]<br /><br />The current ($I$) in a circuit can be calculated using Ohm's law, which states:<br /><br />\[I = \frac{V}{R}\]<br /><br />Substituting the total voltage (9V) and the total resistance (17Ω):<br /><br />\[I = \frac{9V}{17\Omega} \approx 0.53A\]<br /><br />This current is the same throughout the circuit because it is a series circuit. <br /><br />The voltage across each resistor can also be calculated using Ohm's law. For each resistor:<br /><br />\[V = I \times R\]<br /><br />Substituting the current (0.53A) and the resistance for each resistor:<br /><br />For the 2Ω resistor:<br /><br />\[V_1 = 0.53A \times 2\Omega \approx 1.06V\]<br /><br />For the 5Ω resistor:<br /><br />\[V_2 = 0.53A \times 5\Omega \approx 2.65V\]<br /><br />For the 10Ω resistor:<br /><br />\[V_3 = 0.53A \times 10\Omega \approx 5.29V\]<br /><br />#Answer#<br /><br />The completed table is:<br /><br />\begin{array}{|c|c|c|c|}<br\ />\hline\ \begin{array}{c}\ <br\ />Circuit\ \\<br\ />Position<br\ />\end{array} & Voltage (V) & \begin{array}{c}\ <br\ />Current\ \\<br\ />(A)<br\ />\end{array} & \begin{array}{c}\ <br\ />Resistance\ \\<br\ />\ (\Omega)\ <br\ />\end{array} \\<br />\hline 1 & 1.06 & 0.53 & 2.00 \\<br />\hline 2 & 2.65 & 0.53 & 5.00 \\<br />\hline 3 & 5.29 & 0.53 & 10.0 \\<br />\hline Total & 9.00 & 0.53 & 17.0 \\<br />\hline<br />\end{array}