Parallel Circuit en stroomverdeling

- twee elementen zijn parallel als ze tussen hetzelfde paar noten zijn verbonden. Als elk element parallel is met elk ander element, wordt het een parallel circuit genoemd.
- de equivalente weerstand van parallel verbonden weerstand is
  
  $\frac{1}{R_{T}}=\sum_{i}\frac{1}{R_{i}}$
  
  de equivalente geleidbaarheid is de som van de individuele geleidbaarheid

Voorbeeld 1: Zoek de totale weerstandswaarde.,=”c93b939568″>

Voorbeeld 2: Voor de volgende schakeling, de totale resister waarde

de Totale weerstand waarde is

$\frac{1}{R_{T}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}=\frac{1}{1k}+\frac{1}{2k}+\frac{1}{3k}+\frac{1}{4k}=\frac{8}{4k}$

$R_{T}=\frac{4k}{8}=500\Omega$

Voorbeeld 3: Voor de volgende schakeling:

de totale resister waarde RT
het Vinden van de huidige iT
het Vinden van de stroom in elke branch
het Vinden van de macht verdreven door elke weerstand

1., Total resistor value

$\frac{1}{R_{T}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}=\frac{1}{1k}+\frac{1}{2k}+\frac{1}{4k}=\frac{7}{4k}$

$R_{T}=\frac{4k}{7}=571.4\Omega$

2. The total current can be calculated as

$i=\frac{v}{R_{T}}=\frac{12}{571.4}=21mA$

3., The current in each branch

$i_{1}=\frac{12}{R_{1}}=\frac{12}{1k}=12mA$

$i_{2}=\frac{12}{R_{2}}=\frac{12}{2k}=6mA$

$i_{3}=\frac{12}{R_{4}}=\frac{12}{4k}=3mA$

$i_{T}=i_{1}+i_{2}+i_{3}$

Verify that $i_{T}=12+6+3$

$21mA=21mA$

4., The power dissipated by each resistor

$P_{1}=i_{1}V_{1}=12(10^{-3})12=144mW$

or

$P_{1}=i_{1}V_{1}=i_{1}(i_{1}R_{1})=i^{2}_{1}R_{1}=(12(10^{-3}))^{2}(10^{3})=144mW$

$P_{2}=i_{2}V_{1}=6(10^{-3})12=72mW$

or

$P_{2}=i_{2}V_{1}=i_{2}(i_{2}R_{2})=i^{2}_{2}R_{2}=(6(10^{-3}))^{2}2(10^{3})=72mW$

$P_{3}=i_{3}V_{1}=3(10^{-3})12=36mW$

or

$P_{3}=i_{3}V_{1}=i_{3}(i_{3}R_{3})=i^{2}_{3}R_{3}=(3(10^{-3}))^{2}4(10^{3})=36mW$

Example 4: Find the current of i1, i2, i3, and i4 in the following circuit.,6d3c4″>

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}$

$\frac{1}{R_T}=\frac{1}{90}+\frac{1}{60}+\frac{1}{30}+\frac{1}{20}=\frac{2}{180}+\frac{3}{180}+\frac{6}{180}+\frac{9}{180}$

$\frac{1}{R_T}=\frac{20}{180}$

Taking the inverse

$R_T=\frac{180}{20}=9\Omega$

Finding the voltage of the circuit

$V=(I)(R_T)=(20A)(9\Omega)=180V$

Current of each wire using the circuit voltage

$I_1=\frac{V}{R_1}=\frac{180}{90\Omega}=2A$

$I_2=\frac{V}{R_2}=\frac{180}{60\Omega}=3A$

$I_3=\frac{V}{R_3}=\frac{180}{30\Omega}=6A$

$I_4=\frac{V}{R_4}=\frac{180}{20\Omega}=9A$

Current division:

Consider the following circuit., De spanning V daling over elk van de weerstanden in parallel gegeven in termen van stroom en weerstanden.,

$v=i(R_{1}\left\|R_{2}\right\|R_{3}\left\|R_{4})=iR_{eq}$

The current ii through the resistor Ri can be calculated using voltage

$i_{i}=\frac{v}{R_{i}}=\frac{R_{eq}}{R_{i}}i$

Examples:

Example 5: Given the following circuit, find the current i2

Current division

$10mA = 4 mA + i_2$

$i_2 = 10mA - 4mA$

$i_2 = 6mA$

Example 6: For the following circuit, determine i1 to i3.,

The voltage can be calculated as

$\frac{1}{R_T}=\frac{1}{400}+\frac{1}{600}+\frac{1}{480}$

$R_P=160\Omega$

$i_T=6+10=16mA$

$v=IR_T=16mA(160\Omega) =2.56V$

$i_1=\frac{V}{R_1}=\frac{2.56}{400}=6.4mA$

$i_2=\frac{V}{R_2}=\frac{2.56}{600}=4.266mA$

$i_3=\frac{V}{R_3}=\frac{2.56}{480}=5.333mA$

To check result:

$i_T=i_1+i_2+i_3=6.4+4.266+5.333=16mA$

Example 7: For the following circuit, determine i1 to i3.,The voltage can be calculated as

$\frac{1}{R_P}=\frac{1}{400}+\frac{1}{200}+\frac{1}{200}=\frac{5}{400}$

$R_P=\frac{400}{5}=80\Omega$

$R_T=R_1+R_P=320+80=400\Omega$

$i_T=\frac{V}{R_T}=\frac{20}{400}=0.05A$

$v_RI=320*0.05=16V$

$i_1=\frac{V_P}{R_2}=\frac{4}{400}=0.01A$

$i_2=\frac{V_P}{R_3}=\frac{4}{200}=0.02A$

$i_3=\frac{V_P}{R_4}=\frac{4}{200}=0.02A$

To check result:

$i_T=i_1+i_2+i_3=0.01+0.02+0.02=0.05A$

Practice Problems:

(Click image to view solution)

Practice 1: Find the voltage V1, V2 and the current I1, I2 for the following circuit.,3″>

Then $I_1=\frac{V_a}{60}=\frac{2.4}{60}=40mA$

$I_2=\frac{V_a}{240}=\frac{2.4}{240}=10mA$

$I_3=\frac{V_a}{240}=10mA$

So $V_1=I_3\bullet 160=10mA(160)=1.6V$

$V_2=I_3\bullet 80=10mA(80)=0.8V$

Node 1 Voltage as $V_a$

$I_1=\frac{V_a}{60}$

$I_2=\frac{V_a}{240}$

$I_3=\frac{V_a}{160+80}$

Practice 2: Find the current of i1, i2, and i3 through each parallel product.,iv>

The current is $I=400mA$

Then $V_1=100\timesI=100(40mA)=4V$

$V_2=200\timesI=200(40mA)=8V$

$V_3=100\timesI=100(40mA)=4V$

Applying Kirchoff’s Law $V_g-V_1-V_2-V_3=0$

$V_g=4+8+4=16V$

The current $I_1$ is $I_1=\frac{V_2}{300\Omega}=\frac{8}{300}=0.0266A$ or $I_1=26.6mA$

Practice 3: Find V1, V2 voltages on the series parallel circuit., id=”38cbc9975a”>

$I_3=\frac{200}{100+200}\bullet I_1 =\frac{200}{300}(0.06)=0.04A$

The voltage on node 2:

$V_2=-I_2(200)=-(0.02)\times 200=-4V$ or $V_2=-I_3(100)=-(0.04)\times 100=-4V$

By Kirchhoff’s voltage law $V_1-V_2+300(I_1)-40V=0$

$V_1+4+300(0.06)-40V=0$

$V_1=-4-18+40$

$V_1=18V$

Practice 4: Find the voltage Vg and the currents I1 & I3 for the following circuit.,”>

Solution:

Apply Kirchhoff’s current law

$-14mA+I_1+I_2-6mA=0$

$I_1+I_2=8mA$

$\frac{V_g}{4k}+\frac{V_g}{6k}=8mA$

$(\frac{1}{4}+\frac{1}{6})V_g=8mA$

$\frac{5}{12k}V_g=8mA$

$=19.2V$

$I_1=\frac{V_g}{4k}=\frac{19.2}{4k}=48mA$

$I_2=\frac{V_g}{6k}=3.2mA$

Practice 5: Find the current of i1 on the following circuit.,:

View Solution

Solution:

Everything is in parallel

$G_T=\frac{1}{4.5k}+\frac{1}{9k}+\frac{1}{3k}$

$G_T=\frac{6}{9k}s$

$R_T=\frac{1}{G_T}=\frac{9k}{6}=1.5k\Omega$

$I(RT)=(8mA)(1.5k\Omega)=12V$

$I_1=\frac{12V}{4.5k\Omega}=2.66mA$

Exercises:

Answers: