9-22-11 Basic DC Circuits II

# Or or eq 1 the equations for voltage division in

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or or Eq 1. The equations for voltage division in series, for resistor x Eq 2. The equations for current division in parallel, for resistor x Eq 3. Kirchhoff’s Voltage Law Eq 4. Kirchhoff’s Current Law Eq 4. Equivalent resistance for series circuits (top) and parallel circuits (bottom) Procedure I. Designing a Voltage Divider

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For this section, we were given a circuit diagram and asked to find values of and (see Figure 1), and then build it with the resistors we could find. This figure was taken directly from the lab manual given. Our results are stated below Figure 1. Figure 1. The circuit given for part I. Calculating Values for and : 1. 2. 3. With these calculations, we found the correct values of and in order to have the given voltage drops across them. However, with the limited stock of resistors available to us, we had to settle for 5.1 kΩ ± 5%, 3.6 kΩ ± 5%, and 6.8 kΩ ± 5% resistors. Table 2. The voltages measured across each resistor Resistor Given Voltage Drop () Measured Voltage Drop () Percent Difference 5 5.00 0.0% 3.5 3.53 0.86% 6.5 6.61 1.7% II. Designing a Current Divider This part is very much like section I, Designing a Voltage Divider . As such, the format between these two sections will be identical, for consistency’s sake. The same thing was asked of us, determine values of resistors, except for all three resistors in parallel. Figure 2. The circuit given for part II. Calculating Values for Each Resistor: 1. 2. 3. After we found each resistor value, we built the circuit. Luckily, two of our resistors matched with our calculations, the 1.0 kΩ ± 5% and 2.0 kΩ ± 5% resistors. However, we once again had to use a 5.1 kΩ ± 5% resistor in lieu of the 5 kΩ used in the calculation. Table 3. The currents measured across each resistor Resistor Given Current () Measured Current () Percent Difference 2 2.00 0.0%
5 5.13 2.6% 10 10.12 1.2% Question 1. If all the resistors in the current divider are scaled by a factor , how does this affect the current going through them?

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