lab6_ds1_20130312.tex

\subsection{DATA SHEET \#1}
	\begin{enumerate}[A.]
		\item	Circuit \#1: Direct measurements of resistors using handheld BK meter.  The uncertainties on these measurements are primarily from the instrument.  Assume an uncertainty of 0.1\%.\\
		
		$R_1 \pm \Delta R_1 = 29.77 \pm 0.0298 \Omega$\\
		$R_2 \pm \Delta R_2 = 50.04 \pm 0.0500 \Omega$\\
		
		\item Circuit \#1: Direct measurement of current I from power supply.  Assume an uncertainty of 1.0\% for the BK Ammeter values.
		
		$I_{POWER SUPPLY} \pm \Delta I = 0.11 \pm 0.0001 \Omega$\\
		
		$\Delta V_{READING\_POWER\_SUPPLY} = 8.6 V$\\
		
		These two numbers allow you to predict (via calculation using Ohm's Law) the equivalent resistance of whatever is attached to the power supply, imagining whatever is attached as a single resistor.  Make a prediction and do so now (ignore uncertainties):
		
		$R_{EQUIVALENT} = 78.18 \Omega$\\
		
		\item
			\begin{enumerate}[1.]
			
				\item Circuit \#1:  For each resistor, calculate potential changes (voltage) using Ohm's Law predicts for the potential change across each resistance.\\
				
				$\Delta V_{1CALC} \pm uncertainty = (I \pm \Delta I) \times [R_1 \pm \Delta R_1] = 5.5044 \pm  0.055 V$\\ 
				
				$\Delta V_{2CALC} \pm uncertainty = (I \pm \Delta I) \times [R_2 \pm \Delta R_2] = 3.2747 \pm 0.033 V$\\
			
				\item Show how you calculated the uncertainty on the first result above:
				
				$\sqrt{0.001^2 + 0.01^2} \times calculated  \Delta V_{1CALC} = 0.01005 * 5.5044 = 0.055$ \\
					
			\end{enumerate}
			
	\end{enumerate}