GREPhysics.NET: GR9277 #33

GR9277 #33

Problem

GREPhysics.NET Official Solution

\prob{33}
\imgsrc{9277_32)

In the circuit shown above, the resistance are given in ohms and the battery is assumed ideal with emf equal to 3.0 volts.

The voltage across resistor $R_4$ is

0.4V
0.6V
1.2V
1.5V
3.0V

Electromagnetism $\Rightarrow$ }Circuits

One can find the voltage across $R_4$ quite easily. The net resistance of all resistors except $R_1$ is $R_{eq}=((1/R_3+1/R_4)^{-1}+R^5)^{-1}+1/R_2=25 \Omega$ . Kirchhoff's Loop Law then gives $V=I(R_1+R_{eq})\Rightarrow I=3/75A$ .

Now that one knows the current, one trivially finds the voltage across $R_2$ to be $IR_2=1$ V. $I^{'}(R_{34}+R_{5})=1$ , since the resistors are in parallel.

Since $R_{34}=1/R_3+1/R_4=1/60+1/30=20\Omega$ , the current $I^{'}=1/(R_{34}+R_5)=1/50$ .

The voltage across either $R_3$ or $R_4$ is just $1-I^{'}R_5=1-30/50=0.4$ , as in choice (A).

Alternate Solutions

qu 2008-10-13 07:24:52	R1, R2 and R5 are in series just ignoring the R3 and R4 parallel block. Equivalent (R1,R2,R5) is 130 ohm. Current through them is i=v(3)/Req(130)=0.02. Now the parallel block has equivalent rsistance as Req1=(1/R3+1/R4)^-1. Voltage drop across parallel bloc is therefore V=i*Req1(20ohm); which is 0.4 approx. so choice is (A) Reply to this comment
chrisfizzix 2008-10-06 12:11:04	I hate all those annoying reciprocal relations. Here's a way to solve the problem while doing as little annoying circuit manipulations as possible, and instead using some simple relations. Find the equivalent resistance of $R_3$ \|\| $R_4$ : $\frac{1}{R} = \frac{1}{R_3} + \frac{1}{R_4} = 20$ ohms. So, the total resistance of the right branch is 20 + 30 = 50 ohms. Find the equivalent resistance of the whole circuit: this is straightforward if you remember that two equal resistors in parallel give you one resistor of half the value. So, our total resistance is 75 ohms. Find the current: 3.0 volts / 75 ohms = 1/25 amps. Find the voltage drop across $R_1$ : this is easy: $V_1 = I R_1 =$ 2.0 V. Now we know that the voltage across the right branch is 1.0 V, and that half the total current is flowing through the branch as the resistance of the middle and right branches are equal. The voltage drop across $R_5$ is $\frac{1}{2}I R_5 = \frac{1}{25} \times \frac{1}{2} \times 30 = 0.6$ V. So, we know the voltage drop over the top portion of the right branch: 1.0 V - 0.6 V = 0.4V, choice A. Reply to this comment

Comments

infiniteseries
2008-10-29 19:37:59

there seems to be an error with this image. Just a heads up.

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qu
2008-10-13 07:24:52

R1, R2 and R5 are in series just ignoring the R3 and R4 parallel block. Equivalent (R1,R2,R5) is 130 ohm. Current through them is i=v(3)/Req(130)=0.02. Now the parallel block has equivalent rsistance as Req1=(1/R3+1/R4)^-1. Voltage drop across parallel bloc is therefore V=i*Req1(20ohm); which is 0.4 approx. so choice is (A)

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qu
2008-10-13 07:23:55

Reply to this comment

chrisfizzix
2008-10-06 12:11:04

I hate all those annoying reciprocal relations. Here's a way to solve the problem while doing as little annoying circuit manipulations as possible, and instead using some simple relations.

Find the equivalent resistance of $R_3$ || $R_4$ : $\frac{1}{R} = \frac{1}{R_3} + \frac{1}{R_4} = 20$ ohms.

So, the total resistance of the right branch is 20 + 30 = 50 ohms.

Find the equivalent resistance of the whole circuit: this is straightforward if you remember that two equal resistors in parallel give you one resistor of half the value. So, our total resistance is 75 ohms.

Find the current: 3.0 volts / 75 ohms = 1/25 amps.

Find the voltage drop across $R_1$ : this is easy: $V_1 = I R_1 =$ 2.0 V.

Now we know that the voltage across the right branch is 1.0 V, and that half the total current is flowing through the branch as the resistance of the middle and right branches are equal.

The voltage drop across $R_5$ is $\frac{1}{2}I R_5 = \frac{1}{25} \times \frac{1}{2} \times 30 = 0.6$ V. So, we know the voltage drop over the top portion of the right branch: 1.0 V - 0.6 V = 0.4V, choice A.

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energy69
2007-10-14 20:39:53

solution is so much more complicated then it needs to be. \...

sawtooth
2007-11-02 13:35:32

Indeed it is. Btw, doesn't "voltage across" mean the voltage drop?Anyway. Since one has found that the current going through $R_{34}$ is 1/50 A just multiply by $R_{34}=R_3 // R_4=20]Omega$ to find a 0.6V drop (the same in both resistors. You can find the two respective currents and apply for $R_4$ if you dont believe, but in the end you are using the fact that the voltage drop is the same. I dont know why the official solution uses $R_5$ .

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ec2250
2007-08-27 11:05:05

I don't understand why the current across R2 is I. Wouldn't the current have split from I1 to I2 and I3 when the initial current reached the first junction? To clarify, I'm confused at the part where we find the voltage across R2. According to the loop rule, wouldn't it be V - I1R1 - I2R2 = 0?

D8less
2007-10-05 21:27:15

You are right. The author of the above answer mistyped it. The second comment from the bottom mentions there is an error in the result that V2=2 using i1=i2. The correct correction is that i1=2i1 as you said.

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hamood
2007-04-02 13:29:37

combined resistance of R3 and R4 is 20. So both the right arm and the middle arm have the same resistance of 50 ohms. Half the current goes through each, meaning I = 3/150. So drop across R4= drop across R3= IR = 3/150 * 20 = 0.4 V

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bterranova
2006-11-02 09:49:28

typo on the typo

R34^-1=1/20

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bterranova
2006-11-02 09:46:07

Another typo:

R34^-1=(1/30+1/60)^-1

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comorado
2006-10-26 01:23:22

Another mistyp: $V_{R_2}=I R_2 =\frac{3}{75} 50 = \frac{1}{25} 50 = 2 V$ instead of $1V$

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comorado
2006-10-26 01:13:28

Must be $R^{-1}_{eq}=((\frac{1}{R_3}+\frac{1}{R_4})^{-1}+R_5)^{-1}+\frac{1}{R_2}$

Instead of $R_{eq}=((\frac{1}{R_3}+\frac{1}{R_4})^{-1}+R_5)^{-1}+\frac{1}{R_2}$

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