# Barriers Replacement Parallel Arrangement and Example Problems by naufalyodya

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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
- Barrier replacement, R
- Strong currents, I
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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