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Barriers Replacement Parallel Arrangement and Example Problems Parallel Barriers circuit Composition is the arrangement of parallel resistance barriers side by side with the ends meet at the branching point. Some of the obstacles are arranged in parallel can be replaced by a substitute barrier. In a parallel circuit, the electric current from the battery through each lamp. A circuit, where there are several different avenues that can be energized is called a parallel circuit. In the figure, three resistors arranged in parallel and the ends of the three resistors are connected together to a voltage source, so that the flow has three different ways to get through each resistor. What is the current through each resistor? It depends on the resistance each resistor. As an example, namely, the potential difference ie each resistor is 12 V. Current through a resistor is given by I = V / R, so that we can calculate the current through the 48 Ω resistor is I = (12 V) / (48 Ω) = 0.25 A. Strong current through the other two resistors, can be calculated in the same way. What would happen if the resistance 12 Ω is taken from the series? Is the current through the 48 Ω resistance to change? Is the current depends only on the potential difference and constraints? Is this case also the same if done on resistance 24 Ω. Branch of a parallel circuit is independent of one another. Other lamps remain lit, even if one of the lights removed. Total current strength in a parallel circuit is the sum of the current strength of each lane. Total current strength in the circuit it is (0.25 A + 0.5 A + 1 A) = 1.75 A. In a parallel circuit, the opposite, the total resistance is equal to the sum of the reciprocals of each resistance. Barriers to substitute the parallel circuit can be determined by the equation: 1 / R. total = 1/R1 + 1/R2 + 1/R3 Note: This barrier is smaller than the resistance of each resistor of the three resistors connected in parallel in the circuit. The placement of two or more resistors in parallel circuit, always reduce the resistance in the circuit replacement. Barriers decreased as each new resistor adds a new flow lines, and increase the total current due to the potential difference unchanged. To calculate the replacement barrier in a parallel circuit, we first need to know that the total current is the sum of the current through the branch. If IA, IB, and IC is the current through the branch and I is the total current, then I = IA + IB + IC. The potential difference between the ends of each resistor is the same, so the current through each resistor, such as RA can be determined from IA = V / RA. Example 4 Ω resistor, resistor 6 Ω and 12 Ω resistor connected in parallel at the ends of the battery 3 V. What The equivalent series resistance? What is the current in the circuit? Completion Steps What is known? RA = 4 Ω, RB = 6 Ω, and RC = 12 Ω coupled parallel V. source = 3 V What is being asked? - Barrier replacement, R - Strong currents, I Answer a) Determine the third replacement resistor resistance 1=1+1+1 Rpar 4 6 12 1=3+2+1 Rpar 12 12 12 1=6 Rpar 12 Rpar = 12 6 Rpar = 2 Ω b) Determine the current strength in the circuit I=V R I=3A 2 I = 1.5 A Solved 1. Two obstacles, each at 10 Ω and 15 Ω coupled parallel and connected with nine potential difference V. Determine the replacement barriers and strong currents in the circuit! 2. Resistor 12 Ω, 15 Ω and 20 Ω coupled parallel and connected to a potential difference of 12 V. What obstacles successor? What a powerful current through these barriers?

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posted: | 10/1/2012 |

language: | English |

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all about physics educations

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