Finding Resistance in a Parallel Circuit

AI Thread Summary
In a parallel circuit with two resistors, R1 is 10 ohms and the equivalent resistance is 6 ohms. The calculation shows that R2 should equal 15 ohms, as determined by the formula for equivalent resistance in parallel circuits. However, the teacher marked this answer incorrect, leading to confusion. It's noted that the effective resistance in parallel must be less than the smallest resistor, reinforcing that 15 ohms is indeed valid. The discussion raises questions about potential additional factors, like the ammeter's shunt resistance, that might affect the answer.
Mashfeek
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So there is a parallel circuit with two resistors, R1 and R2, and an ammeter that are connected to a constant 30-volt source. The equivalent resistance of the circuit is 6.0-ohms. The resistance of R1 is 10.-ohms. The resistance of R2 is equal to
a) 6.0 Ω
b) 2.0 Ω
c) 15 Ω
d) 4.0 Ω



1/Req = 1/R1 + 1/R2



1/6Ω = 1/10Ω + 1/R2, solve for R2
I plugged in the values and I got 15 Ω as my answer, but my teacher marked it as wrong in my homework.
 
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Mashfeek said:
So there is a parallel circuit with two resistors, R1 and R2, and an ammeter that are connected to a constant 30-volt source. The equivalent resistance of the circuit is 6.0-ohms. The resistance of R1 is 10.-ohms. The resistance of R2 is equal to
a) 6.0 Ω
b) 2.0 Ω
c) 15 Ω
d) 4.0 Ω



1/Req = 1/R1 + 1/R2



1/6Ω = 1/10Ω + 1/R2, solve for R2
I plugged in the values and I got 15 Ω as my answer, but my teacher marked it as wrong in my homework.

Welcome to the PF.

I get 15 [STRIKE](well 14.99999...)[/STRIKE] too. Is there more to the question? Does the ammeter have some shunt resistance that you need to take into account?
 
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Mashfeek said:
So there is a parallel circuit with two resistors, R1 and R2, and an ammeter that are connected to a constant 30-volt source. The equivalent resistance of the circuit is 6.0-ohms. The resistance of R1 is 10.-ohms. The resistance of R2 is equal to
a) 6.0 Ω
b) 2.0 Ω
c) 15 Ω
d) 4.0 Ω

1/Req = 1/R1 + 1/R2
1/6Ω = 1/10Ω + 1/R2, solve for R2
I plugged in the values and I got 15 Ω as my answer, but my teacher marked it as wrong in my homework.

When resistors are connected in parallel, the effective resistance is always smaller than the smallest of the individual resistors. That means both of these two resistors has to have a resistance in excess of 6Ω.

The ONLY option to meet those conditions is 15Ω.

Your accurate calculation shows that to be the correct answer.

It appears your teacher has not recognised your correct answer.

EDIT: In case your teacher is being very technical ... Did you say R2 was equal to 15 Ohm, or did you say (c)
 
Are you sure there is nothing special about the circuit? Why are voltage source and the ammeter mentioned?

If the circuit is just two parallel resistors, your result is correct because 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6.
 
berkeman said:
Welcome to the PF.

I get 15 (well 14.99999...) too. Is there more to the question? Does the ammeter have some shunt resistance that you need to take into account?

Indeed: where is the ammeter connected in this circuit?
 
The other values given in the question that aren't useful are meant for other questions. I just really wanted to know if 15-ohms was the answer. My physics teacher makes a lot of mistakes. Thank you guys!
 
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