Parallelschaltung und Stromteilung

- Zwei Elemente sind parallel, wenn sie zwischen demselben Notenpaar verbunden sind. Wenn jedes Element parallel zu jedem anderen Element ist, wird es als Parallelschaltung bezeichnet.
- Der äquivalente Widerstand des parallel geschalteten Widerstands ist
  
  $\frac{1}{R_{T}}=\sum_{i}\frac{1}{R_{i}}$
  
  Die äquivalente Leitfähigkeit ist die Summe der einzelnen Leitfähigkeit

Beispiel 1: Finden Sie den Gesamtwiderstandswert.,=“c93b939568″>

Beispiel 2: Für die folgenden schaltung, finden die insgesamt resister wert

Insgesamt widerstand wert ist

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

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

Beispiel 3: Für die folgende Schaltung:

Finden Sie den Gesamtresistorwert RT
Finden Sie den Strom IT
Finden Sie den Strom in jedem Zweig
Finden Sie die von jedem Widerstand abgegebene Leistung

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., Die Spannung v fällt über jeden der Widerstände parallel in Bezug auf Strom und Widerstände gegeben.,

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

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: