Unknown value of added resistor in parallel

In summary: After that, I'll bet something will occur to you. :)In summary, the unknown resistance R in the parallel circuit is 5.2 ohm, while the unknown resistance R in the series circuit is 3.8 ohm.
  • #1
grantaere
15
0

Homework Statement


Picture provided.
The current going through the resistor R1 in the figure does not change whether the two switches S1 and S2 are both open or both closed.
The resistances are R1 = 14.00 Ω, R2 = 142.00 Ω, and R3 = 24.00 Ω. The voltage is V = 11 V.
With this information, what is the value of the unknown resistance R? Recall that a conducting wire can be treated as a resistor of 0 Ω, and points on a wire between resistors are at the same potential.

Homework Equations


parallel circuit: 1/R = 1/(1/R1 + 1/R2)
series circuit: R = R1+R2
parallel: Vtotal = V1 = V2
series: Vtotal = V1 + V2
V = IR

The Attempt at a Solution


I tried finding the total resistance of the circut when switches are open, then using that with total voltage (11V) to find the current that goes through each resistor in series. (0.06111A) Then I used V = IR to find the voltage drops across each resistor in the series circuit. (V1 = 0.8555, V2=8.67777, V3=1.4666)
Then, I tried finding the resistance of R + R1 when the switches are closed-- R in parallel with R1-- and setting that equal to Vtotal/total current going through the juncture... but it gives me zero as an answer which is obviously wrong. Any hints would be very appreciated! I'm also quite confused as to how the outside switch factors into the problem at all.
 
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  • #2
You need to draw the circuit with the switches both open and write the loop equations, then draw the circuit with the switches both closed and write the loop equations. After that, I'll bet something will occur to you. :)
 
Last edited:
  • #3
you already know that I1=0.06666A and V1=0.8555V ... the only other voltage (closed switches) is V2.
(once you determine a V, you can find a new I)
 
  • #4
Okay, so I tried drawing the second circuit (with switches closed) and found that R3 becomes zero since it's in parallel with a wire (assuming this is true?). In this case, I tried making the new total voltage = 11 = IR(2) + IR(unknown + 1 in parallel). Since it says the current remains the same across the R1 juncture regardless of whether the switches are closed, I assumed the current remains 0.06111A, and found the R in parallel to equal 3.8 ohm. Plugging that into 1/(1/R1 + 1/R) = R(parallel) I got the unknown R to equal 5.2ohm... but apparently this is still incorrect. What am I doing wrong?
 
  • #5
I suggest, again, that you do what I suggested in post #2
 

1. What is the unknown value of the added resistor in parallel?

The unknown value of the added resistor in parallel is the resistance of the resistor that is connected in parallel with other resistors in a circuit. It is denoted by the symbol "Rx" and is usually calculated using the formula: 1/Rx = 1/R1 + 1/R2 + ... + 1/Rn, where R1, R2, ..., Rn are the known resistances in the circuit.

2. Why is the unknown value of the added resistor in parallel important?

The unknown value of the added resistor in parallel is important because it affects the overall resistance and current in the circuit. By knowing the value of Rx, we can calculate the total resistance and current in the circuit, which is crucial for designing and analyzing electrical circuits.

3. How can the unknown value of the added resistor in parallel be measured?

The unknown value of the added resistor in parallel can be measured using a multimeter, which is a device that measures electrical properties such as resistance, voltage, and current. By connecting the multimeter in parallel with the unknown resistor, we can determine its resistance value.

4. What happens if the value of the added resistor in parallel is too low?

If the value of the added resistor in parallel is too low, it will cause an increase in current flow through the circuit. This can lead to overheating and potential damage to the circuit components. It is important to choose the correct value for the added resistor to prevent any potential issues.

5. What happens if the value of the added resistor in parallel is too high?

If the value of the added resistor in parallel is too high, it will cause a decrease in current flow through the circuit. This can affect the functionality of the circuit and may result in inaccurate readings or improper functioning of electrical devices. It is important to choose the correct value for the added resistor to ensure the proper functioning of the circuit.

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