Question

HW Practice #1: Parallel Circuits $I = I_1 + I_2 + I_3 + ...$ $V = V_1 = V_2 = V_3 = ...$ $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...$ $R = \frac{V}{I}$ $P = VI = I^2R = \frac{V^2}{R}$ + + + = Parallel P R I V R1 R2 R3 T 9V + - $R_1$ 90 ? $R_2$ 45 ? $R_3$ 180 ?

          HW Practice #1:
Parallel Circuits
$I = I_1 + I_2 + I_3 + ...$
$V = V_1 = V_2 = V_3 = ...$
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...$
$R = \frac{V}{I}$
$P = VI = I^2R = \frac{V^2}{R}$
+     +     +     =     Parallel
P     R     I     V
R1
R2
R3
T
9V
+     -
$R_1$
90 ?
$R_2$
45 ?
$R_3$
180 ?

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HW Practice #1:
Parallel Circuits
I = I1 + I2 + I3 + ...
V = V1 = V2 = V3 = ...
(1)/(Req) = (1)/(R1) + (1)/(R2) + (1)/(R3) + ...
R = (V)/(I)
P = VI = I^2R = (V^2)/(R)
+ + + = Parallel
P R I V
R1
R2
R3
T
9V
+ -
R1
90 ?
R2
45 ?
R3
180 ?

Added by Megan A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Krishna G

02/15/2022

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This can be represented as I = I1 + I2 + I3 + ... Show more…

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HW Practice #1: Parallel Circuits I=I_(1)+I_(2)+I_(3)+dots V=V_(1)=V_(2)=V_(3)=dots (1)/(R_(eq))=(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))+dots R=(V)/(I) P=VI=I^(2)R=(V^(2))/(R) P=VI=I2R= R= V=V=V=V-.. I=1+1+1+. Parallel Circuits +R HW Practice #1 +.. + + = V Parallel R R1 4HP oV R2 90 ER, 45 180 R3

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Transcript

00:01 In this problem a circuit is given and we have to find out the current that is i1.

00:09 Okay, so our circuit is.

00:11 So this is our circuit.

00:12 We have i that is current and current is passing through in this direction.

00:19 And 24 volt is voltage and this is our resistance.

00:25 100 om, 250 om and 200 om.

00:28 And we can see that 100 om and 100 om.

00:31 And 250 oom are in parallel combination, this and this, okay? and 350 and 200 is also in parallel combination.

00:41 Okay.

00:42 And this hole and this hole is in series combination.

00:48 Okay.

00:49 So suppose this is first resistance.

00:51 This is second resistance.

00:54 And this is third resistance and this is fourth resistance.

00:59 So here we have to find out the current, okay, i1.

01:04 So first of all, we can see that 100 oom and 250 oom are in parallel combination.

01:10 So we know that the formula of parallel combination for finding resistance is 1 by rp is equals to 1 by r1 plus 1 by r2.

01:25 So this is our formula.

01:26 Okay, so suppose we have to find 100 oom and 250 ome resistance.

01:33 So 1 by r 1 2, 1 for 100 om and 2 for 250 om.

01:40 That is equals to 1 by r1 that is 100 plus 1 by r2 that is 250.

01:48 These are in parallel combination.

01:56 Parallel combination...

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