# Confused by this circuit problem

#### acetylmandarin

2+ Year Member
Why wouldn't both Opening switch A and opening switch B affect the power drain on resistor III? If you opened switch A, wouldn't that mean that all of the current is now diverted to resistor III (thus changing the power because P = I^2 (R) and if you opened switch B, wouldn't that also change the power because that would mean 0 current is flowing through the circuit?

In the answer explanation, they say that the equivalent resistance would change, but the current and voltage drop through resistor III would be unaffected. How can you change the resistance without changing the current? I tried to draw it out and use numbers to figure it out but I got really lost.

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#### bobeanie95

2+ Year Member
If you opened swtich A all the current would flow through only through resistor III, since the number of resistors the current is passing through changes, so would the Requivalent and the power emitted. In all the other answer choices, the modifications would produce an incomplete circuit so current won't be able to flow.

OP

#### acetylmandarin

2+ Year Member
If you opened swtich A all the current would flow through only through resistor III, since the number of resistors the current is passing through changes, so would the Requivalent and the power emitted. In all the other answer choices, the modifications would produce an incomplete circuit so current won't be able to flow.

So yeah, that's what I thought. I thought the power drain would change for both choices. But they say the answer is B. Somehow, when opening switch A, the resistance changes without the current or voltage changing

#### bobeanie95

2+ Year Member
Well B says its opening switch B not A. I figured the answer would be A.

OP

#### acetylmandarin

2+ Year Member
Well B says its opening switch B not A. I figured the answer would be A.
Yeah, I don't understand. I see both as changing the power

#### bobeanie95

2+ Year Member
Yeah, I don't understand. I see both as changing the power
It may just be a bad question. I don't see how the answer could be B, because opening the B switch would make an open circuit even if A remains closed.

OP

#### acetylmandarin

2+ Year Member
It may just be a bad question. I don't see how the answer could be B, because opening the B switch would make an open circuit even if A remains closed.
The question is asking which would alter the power drain. So wouldn't both apply then? Since if you opened the circuit the power drain would go to 0

#### aurevoir0711

5+ Year Member
I was really confused on this question as well, but I figured TBR questions usually do not contain logic errors, so I spent some time to understand what the heck is going on.

If you actually work out the math, you will see that voltage drop remains as 12V and current remains as 2A for resistor III in both cases (switch closed or open). Since P = IV, dissipated power in both cases are identical.

I think the key takeaway from this question is that resistor in series "consumes" voltage while parallel circuits "preserves" voltage drop. Whatever the voltage difference before the parallel circuit will be the voltage drop across each resistor in parallel circuit. In series circuits, current remains same throughout (as we know), and for voltage, whatever the voltage drop before entering would be subtracted by however much resistance the series circuit has. This question really reinforced the concept about V1 = V2 = V3 in parallel circuits.

#### Pono_0001

ya you need to understand that resistance in parallel always equate to the voltage of the battery. according to Ohm's law, that means the current flowing through a resister in parallel is always the same, unless the R or the V changes. That always means total current would change when total R in parallel changes. so your P = I^2 (R) holds describing Total R, but since current flowing through RIII doesn't change, P stays the same. Only opening B would change I flowing through RIII