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Using Kirchhoff's rules, find the following. ($\mathcal{E}_1 = 70.4$ V, $\mathcal{E}_2 = 62.0$ V, and $\mathcal{E}_3 = 78.4$ V.) b c 4.00 k$\Omega$ d $\mathcal{E}_1$ $\mathcal{E}_2$ R$_3$ $\mathcal{E}_3$ R$_2$ 3.00 k$\Omega$ 2.00 k$\Omega$ a R$_1$ f e (a) the current (in mA) in each resistor shown in the figure above I$_{R_1}$ = x Apply Kirchhoff's rules to write a set of simultaneous equations that can be solved for the current in each part of the circuit. mA I$_{R_2}$ = x Apply Kirchhoff's rules to write a set of simultaneous equations that can be solved for the current in each part of the circuit. mA I$_{R_3}$ = x Enter a number. f's rules to write a set of simultaneous equations that can be solved for the current in each part of the circuit. mA

          Using Kirchhoff's rules, find the following. ($\mathcal{E}_1 = 70.4$ V, $\mathcal{E}_2 = 62.0$ V, and $\mathcal{E}_3 = 78.4$ V.)
b c 4.00 k$\Omega$ d
$\mathcal{E}_1$ $\mathcal{E}_2$ R$_3$ $\mathcal{E}_3$
R$_2$ 3.00 k$\Omega$
2.00 k$\Omega$
a R$_1$ f e
(a) the current (in mA) in each resistor shown in the figure above
I$_{R_1}$ = x
Apply Kirchhoff's rules to write a set of simultaneous equations that can be solved for the current in each part of the circuit. mA
I$_{R_2}$ = x
Apply Kirchhoff's rules to write a set of simultaneous equations that can be solved for the current in each part of the circuit. mA
I$_{R_3}$ = x
Enter a number. f's rules to write a set of simultaneous equations that can be solved for the current in each part of the circuit. mA

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using kirchhoffs rules find the following mathcale1 704 v mathcale2 620 v and mathcale3 784 v b c 400 komega d mathcale1 mathcale2 r3 mathcale3 r2 300 komega 200 komega a r1 f e a t 58922

Added by Donna W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Mark Scythian

06/25/2025

Step 1

4$ V $\mathcal{E}_2 = 62.0$ V $\mathcal{E}_3 = 78.4$ V $R_1 = 2.00$ k$\Omega = 2000 \Omega$ $R_2 = 3.00$ k$\Omega = 3000 \Omega$ $R_3 = 4.00$ k$\Omega = 4000 \Omega$ We need to find the currents $I_{R_1}$, $I_{R_2}$, and $I_{R_3}$. We will use Kirchhoff's Show more…

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Using Kirchhoff's rules, find the following. ($\mathcal{E}_1 = 70.4$ V, $\mathcal{E}_2 = 62.0$ V, and $\mathcal{E}_3 = 78.4$ V.) b c 4.00 k$\Omega$ d $\mathcal{E}_1$ $\mathcal{E}_2$ R$_3$ $\mathcal{E}_3$ R$_2$ 3.00 k$\Omega$ 2.00 k$\Omega$ a R$_1$ f e (a) the current (in mA) in each resistor shown in the figure above I$_{R_1}$ = x Apply Kirchhoff's rules to write a set of simultaneous equations that can be solved for the current in each part of the circuit. mA I$_{R_2}$ = x Apply Kirchhoff's rules to write a set of simultaneous equations that can be solved for the current in each part of the circuit. mA I$_{R_3}$ = x Enter a number. f's rules to write a set of simultaneous equations that can be solved for the current in each part of the circuit. mA

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00:01 So we can start by solving current in each resistor.

00:10 So we can define i1 as current through r1, left to right, i2 as current through r2 upward, and i3 current through r3 downward.

00:23 So using kirchhoff's current law, kcl, we have the first equation to implement here, and that is i1 is equal to i2 plus i3 so call that equation one second is the kerkhov's voltage law so kvl rather than kerkov's current law kcel so we can start with the left loop and this is clockwise so this is described as negative voltage one source plus i1 times r1 plus i2 times r2 minus voltage 2 source is equal to 0 yields 15 i1 plus 25 i2 is equal to 132 .2.

02:18 So we'll call this equation 2.

02:24 Then we have the right loop.

02:38 This is also clockwise.

02:48 So we have negative voltage 3 plus i3 times r3 minus i2 times r2 plus voltage source 2 is equal to 0.

03:26 3 minus 25 i2 is equal to 17 .8.

03:38 So this is equation 3.

03:44 So next we solve the system so we take equation 1 which is the i 1 is equal to i 2 plus i 3 and we substitute it into equation 2.

03:55 So we'll get 15 distributed across i 2 plus i 3 plus 25 i is equal to 132 .2, which will yield 40 i .2 plus 15 i3 equal to 132.

04:42 So we call this equation 4.

04:47 So then i3 is 25 i2 plus 17.

05:03 Divided by 20.

05:09 So now we substitute this into equation number four and this will then become 40 i2 plus 15 times the ratio of 25 i2 plus 17 times the ratio of 25 i 2 plus 17 .8 divided into 20.

05:55 So this will then, is made equal to 132 .2, which will then yield 40 i2 plus 18 .75, i2 plus 13 .35.

06:31 13 .35 is equal to 1332 .2...

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