Pandora32
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can anyone help me with this question. I don't understand how to find the answer. Thanks
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uberteknik
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(Original post by Pandora32)
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can anyone help me with this question. I don't understand how to find the answer. Thanks
This is an application of both Ohms and Kirchoff's laws.

Notice that there are two paths current can take at the junction between the 1k and 2k Ohms resistors?

We can ignore the 1k path as the current in this path flows directly back to the supply and has no affect on the potential difference between nodes X and Z.

The current flowing in the 3k resistor places this resistor in series with the 2k resistor.

i.e. the total resistance in the series current path is 2k + 3k = 5k Ohms.

Using Ohms law:

I = V/R = 100/5000 = 0.02 Amps

from KCL (Kirchoffs Current Laws), the series path current must be the same at all points in the current path.

i.e. 0.02A must flow through the 3k resistor and once again from Ohms law the p.d. developed across the 3k resistor is:

V = I x R = 0.02 x 3000 = 60V.
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