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The meter in the circuit shown in the image has an uncalibrated linear scale. With the circuit as shown, the scale reading is 20. Find the scale reading when another 2000Ω resistor is connected across XY.
i agree with you.Work out the currents for each setup. Initially, all components are in series.
1) 1.5 volts / ( 3000 ohms + 2000 ohms) =
Now, 2000 ohms in parallel with 2000 ohms = 1000 ohms.
2) 1.5 volts / ( 3000 ohms + 1000 ohms ) =
Then, if the first current = "20" on the linear scale, what does the second current read?
It will read more, won't it?
This is exactly what franklinear showed. Neither of you showed how you came up with 1000 ohms though.No.
Work out the two values with your calculator.
What does this equal:
1) 1.5 volts / ( 3000 ohms + 2000 ohms) =
Add 3000 and 2000 and divide 1.5 by this total.
What does this equal:
2) 1.5 volts / ( 3000 ohms + 1000 ohms ) =
Add 3000 and 1000 and divide 1.5 by this total.
You should get two currents in Amps.
Now the meter reading will be 20 times the ratio of (current 2 / current 1)
I agree with your answer but I don't think you know how you got it.
This is clearly course work, so we can't just do it for you.
Hey vk6kro, is the total resistance of parallel resistors equal to the highest resistance resistor divided by the number of resistors in parallel?
I am not a physics major or anything but I do like to toy around with equations. I am sure this is a basic question in your realm.
This is completely true (and it is derived from the formula I posted). The reason I didn't mention it, is that this formula only works for 2 resistors in parallel at a time, whereas the one I gave can be extended to any number of resistors in parallel, by:Another way of getting the total of resistors in parallel, especially if they are not the same, is this:
Rtotal = R1 * R2 / (R1 + R2)