91Ó°ÊÓ

First, we have to find out the current that passes through Oscar while holding the conducting wires. We can use Ohm's Law to calculate the current passing through Oscar: \(I = \frac{V}{R}\), where \(V = 100.0 V\) and \(R = 2.0 k\Omega\). Converting \(R\) to \(\Omega\), we get \(R = 2000\Omega\). Now, we can find the current: \(I = \frac{100.0}{2000} = 0.05 A\). So, \(0.05 A\) current passes through Oscar.

In this case, we are told that Oscar and a \(15\Omega\) resistor are connected in parallel. To calculate the current passing through Oscar, we first have to find the equivalent resistance of the parallel connection using the formula: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\), where \(R_{eq}\) is the equivalent resistance, \(R_1 = 2000\Omega\) (Oscar's resistance) and, \(R_2 = 15\Omega\). Now, we substitute the given values into the equation: \(\frac{1}{R_{eq}} = \frac{1}{2000} + \frac{1}{15}\). Solving this equation, we get that \(R_{eq} = 14.77\Omega\). Since the maximum current that can be drawn from the emf is \(1.00 A\), we need to find out the voltage drop across the equivalent resistor. Using Ohm's Law: \(V_{eq} = IR_{eq}\), where \(I = 1.00 A\) and \(R_{eq} = 14.77\Omega\). Substituting the values, we get \(V_{eq} = 1.00 \times 14.77 = 14.77 V\). Now, we can calculate the current passing through Oscar using Ohm's Law: \(I_O = \frac{V_{eq}}{R_O}\), where \(V_{eq} = 14.77 V\) and \(R_O = 2000\Omega\). Substituting the values, we get \(I_O = \frac{14.77}{2000} = 0.007385 A\). So, when one of the conducting wires is grounded and the other has an alternate path to the ground through a \(15\Omega\) resistor, \(0.007385 A\) current would pass through Oscar.