In the physics laboratory, Oscar measured the resistance between his hands to be \(2.0 \mathrm{k} \Omega .\) Being curious by nature, he then took hold of two conducting wires that were connected to the terminals of an emf with a terminal voltage of \(100.0 \mathrm{V} .\) (a) What current passes through Oscar? (b) If one of the conducting wires is grounded and the other has an alternate path to ground through a \(15-\Omega\) resistor (so that Oscar and the resistor are in parallel), how much current would pass through Oscar if the maximum current that can be drawn from the emf is \(1.00 \mathrm{A} ?\)

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.