Circuit with Resistors and a Battery

AI Thread Summary
The discussion focuses on a circuit with five resistors and a 12 V battery, where the current through resistor R4 is zero. This condition implies that the voltage across R4 is also zero, indicating equal potential at both ends. Participants express confusion about calculating voltages and currents across the resistors, particularly R2 and the unknown resistor RX. The relationship between the resistors in series and the concept of voltage division is highlighted, emphasizing that the voltage across each resistor depends on their respective values. Overall, the thread seeks clarity on applying Ohm's law and understanding the implications of zero current in the circuit.
hawaldko
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Homework Statement


A circuit is constructed with five resistors and a battery as shown. The battery voltage is V = 12 V. The values for the resistors are: R1 = 54 Ω, R2 = 142 Ω, R3 = 148 Ω, and R4 = 109 Ω. The value for RX is unknown, but it is known that I4, the current that flows through resistor R4, is zero.

http://imgur.com/IwvnK

http://imgur.com/IwvnK

There is a picture of the circuit.

I know the current through R1 is 0.0594059406.

Here are my questions:

1) What is V2, the magnitude of the voltage across the resistor R2?
2)What is I2, the magnitude of the current that flows through the resistor R2?
3)What is RX, the value of the unknown resistor RX?
4)What is V1, the magnitude of the voltage across the resistor R1?


Homework Equations



I=V/R

The Attempt at a Solution



1) Essentially, I thought the voltage across V2 should be 12, the same as the battery because the voltage should be equal across parallel circuits. This answer is wrong, and I don't really understand why. I don't know how to use I=VR without knowing the current, which I know won't be the same and not knowing Rx.

2) I think I need the answer from 1 to do this one, and I can just use V=IR, but I don't know how to do it was Rx instead of a real number

3) Absolutely no idea.

4) I think this should be 12, but I don't think it is because of the first problem.

I am really having difficulty with this problem and would appreciate any help.
 
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If the current through R4 is zero, then all the current that flows through R1 also flows through R3. Similarly, all the current that flows through R2 must flow through Rx.

If the current through R4 is zero, what does that say about the potential at each end of R4?
 
So the potential has to be the same? How can it be the same if there are different resistors before it? I'm sorry, I'm just really not getting it.
 
hawaldko said:
So the potential has to be the same? How can it be the same if there are different resistors before it? I'm sorry, I'm just really not getting it.

The current is zero so there cannot be a difference in potential. So you need to look at how the potential at either end of that resistor is established. Consider R1 and R3. Two resistors in series form a voltage divider. How does the voltage divide across the two resistors?
 
So does that mean the the voltage across each of them is 6?
 
hawaldko said:
So does that mean the the voltage across each of them is 6?

No. How does it depend upon the resistor values? Can you work it out?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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