Practice problem: Determining R using current divider method

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SUMMARY

The discussion centers on a practice problem involving the current divider method to determine the value of a load resistor (R_L). The user calculated R_L to be 8 ohms based on the currents through a 10-ohm resistor and R_L, which were 0.2 A and 0.25 A, respectively. However, the textbook states the correct answer is 24 ohms. After verifying with the Multisim program, the user remains confident in their calculation, leading to confusion regarding the textbook's answer.

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Saeed.z
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Hi all ,

there is a practice problem in my textbook which say :

http://im17.gulfup.com/2012-02-14/1329248560881.jpg

i found after the load is attached that the current through 10 ohm's is 0.2 ampere and through R_L is 0.25 ampere, therefore , the value of R_L is 8 ohm's but the correct answer is 24 ohm's !

i checked again with the multism program and found my answer is correct !

i used the current divider method here .

so I'm a little confused ! which one is correct ?

BTW , sorry for my bad english :)

thanks

 
Last edited by a moderator:
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Saeed.z said:
Hi all ,

there is a practice problem in my textbook which say :

http://im17.gulfup.com/2012-02-14/1329248560881.jpg

i found after the load is attached that the current through 10 ohm's is 0.2 ampere and through R_L is 0.25 ampere, therefore , the value of R_L is 8 ohm's but the correct answer is 24 ohm's !

i checked again with the multism program and found my answer is correct !

i used the current divider method here .

so I'm a little confused ! which one is correct ?

BTW , sorry for my bad english :)

thanks


The current drawn from the source power supply is greater when the Rl resistor is connected.

To figure this out, just write the equation for the output voltage with the known resistors 40 and 10 and the unknown Rl in parallel with the 10 Ohm resistor.
 
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I get 8 Ohms too.
 
DragonPetter said:
I get 8 Ohms too.

Interesting. I didn't even bother to figure out the answer, but you are right. Plugging in 8 Ohms for the load resistor does give 2V at the output. Hmm...
 
berkeman said:
Interesting. I didn't even bother to figure out the answer, but you are right. Plugging in 8 Ohms for the load resistor does give 2V at the output. Hmm...

I tried to figure out where they could get 24 ohms from, as if I misinterpreted the problem, but I didn't bother with spending the time and assume its an error in the answer book.
 
@berkeman
@DragonPetter
thank you all of you guys , i really appreciate your help !
thanks !
 

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