If the switch in the circuit is closed for a very long time, what is the charge on the capacitor? (R1 = 1274 ,?

It's not possible in the real world. When the switch is open the current path is R1, R2, R3, which is the most resistance the battery will ever see, so it should be the lowest current. In position a, the R2 resistors are placed in parallel, and must have less total resistance than either.

In this case, it's one half since they are the same value. This lower total resistance should net an INCREASE in current, not a decrease.

There's something wrong with your values which are inconsistent. Here's the method though - when you repeat it with the correct values you will get the correct answer. With the switch open, the current only flows through the outer 3 resistors, which act in series.

Total resistance = R1 + R2 +R3. Using V=IR: 6 = 0.0018 x (R1+R2+R2) R1+R2+R3 = 3333 (equation 1) With the switch in position a, the two R2s are in parallel, with an equivalent resistance of ½R2. The '½R2' is in series with R1 and R3 giving a total resistance = R1+½R2+R3 Using V=IR 6 = 0.0014 x (R1+½R2+R3) R1+½R2+R3 = 4286 (equation 2) With the switch in position b, R3 is short-circuited - which means the current takes the 'easy' route and doesn't pass through R3.

So the current only flows through R1 and the outer R2. Total resistance = R1+R2. Using V=IR 6 = 0.0023 x (R1+R2) R1+R2 =2609 (equation 3) Subtract equation 3 from equation 1 R3 = 3333-2609 = 724?

= 720? To 2 significant figures Subtract equation 3 from equation 2 ½R2 = 4286-2609 = 1677 R2 = 3354? = 3400?

To 2 significant figures From equation 3 R1 = 2609 - 3354 = -745? = -75? To 2 significant figures of course the resistance can't be negative, but, as I said, you have made a mistake in the question.

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