Voltage/Circuits Statements False?

[A_lilah]

Homework Statement

A circuit consists of 2 resistors, R1 and R2, connected in parallel across a battery.
Which statement is false?

(A) The greater power is produced in the smaller resistor.
(B) If you place another resistor in series with R1 and R2, you will
increase the overall resistance of the circuit.
(C) If you place another resistor in parallel with R1 and R2, you will
decrease the overall resistance of the circuit.
(D) If the battery is ideal, and you disconnect R2ʼs branch from the
circuit, the power dissipated in R1 will not change.
(E) If the battery is non-ideal, and you disconnect R2ʼs branch from the
circuit, the voltage difference across R1 will decrease.
(F) None of the above.

Homework Equations

P = ∆VI = (∆V^2)/R = (I^2)R
Series: R total = R1 + R2 + R3...
Parallel: 1/ R total = 1/R1 + 1/R2 + 1/R3...

The Attempt at a Solution

(A) True because P = (∆V^2)/R, and the smaller the resistor, the smaller the denominator, making the power larger.
(B)True because resistors in series are additive (R1+R2+R3... = R total), so adding one in series would just increase the circuit resistance
(C) True. I had to run some math to check this one, so I arbitrarily chose R1 = 2, R2 = 3, and the added R3 = 4:

Without the added resistor (R1 + R2)-
1/(R total) = 1/2 + 1/3 = 5/6, R total = 6/5, or 1.2 ohms

With the added resistor (R1+ R2 + R3)-
1/ (R total) = 1/2 + 1/3 + 1/4 = 13/12, R total = 12/13, or 0.92 ohms, which is obviously less than above


This is where I got confused:
(D) If you disconnect R2, then the ∆V across R1 will be greater, making its power greater, so I thought this one was false.

(E) If you disconnect R2, then the voltage difference across R1 will increase, right? That was my logic for (D) anyway... I'm not entirely sure how the battery being non-ideal fits in, but I think that for simplicity's sake, its like having another resistor (the 'internal resistance' resistor, in series with R1), and by removing R2, you are increasing the resistance across both which would increase the voltage change for both...

Any input would be great. Thanks!

 

[skeptic2]

Let's look at (D) and (E) where you admit to some confusion. The ideal battery means it has no internal resistance so the battery would be capable of infinite current at the rated voltage. If it were not ideal there is some internal resistance that you must include in your calculations.

So for (D) with no internal resistance, what is the voltage across R1 with R2 in the circuit and what is it without R2 in the circuit?

For (E) with internal resistance, which is in series with the combination of R1 and R2, what happens to the voltage across R1 when R2 is removed.

Don't forget (F). This problem is badly written because if any of the other answers are false, (F) is also false and so would have to be included in the answer. If the answer to none of the other questions is false then (F) isn't false either so there would be no "correct" answer.

 

[xcvxcvvc]

For (D), the voltage across the resistor does not change if you remove the parallel component, so using the equation for power you provided to answer (A), we see that voltage and resistance remain the same. Therefore, the power too remains the same.

For (E), you can use voltage division. That is if n resistors are in series, the voltage across some resistor k equals(where k's range is [1, n]):
V_{k} = V_{Battery}\frac{R_{k}}{R_{1} + ... R_{n}}
So as you answered correctly in (C), adding resistors in parallel decreases resistance. Sensibly, removing resistors in parallel increases resistance. As we see in the equation for voltage division, the voltage across R_k increases as its contribution to the total series resistance increases. It also decreases as its contribution decreases. Therefore, removing a resistor increases the resistance, increasing the voltage across R_1

 

[A_lilah]

Oohhhh... Thanks! I guess it makes sense that removing a resistor from a parallel circuit does not change the voltage. :)