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    The equation for voltage at a point, V1, across a resistor with R1 resistance in a potential divider with resistors R1 and R2 is given by:

    V_1 = V_o + \Delta V \frac{R_1}{R_1 + R_2}, where \Delta V is the voltage across the potential divider and V_o is the voltage at reference point.

    Here is the problem I am solving:



    The question is to find the voltage V3 and the given answer is -10 V.

    The arrow is across the 60 \Omega resistor, so I calculated:

    V_3 = -14 + [-2 - (-14)] \times [\frac{60}{60 + 30}] = -6 V, which is not the right answer.

    I get -10 V if I replace the 60 ohms on the numerator with 30 ohms, which is the resistance of the other resistor.

    V_3 = -14 + [-2 - (-14)] \times [\frac{30}{60 + 30}] = -10 V

    Could you explain why is that the case?
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    It's east to see, either using the potential divider equation (or more quickly by noting the ratio of the resistors) that there is 8 V across the 60 ohm resistor and 4 V across the 30 ohm resistor. The only confusion is what the question is actually asking! I think it is trying to get you to say what the potential is at a point between the resistors, taking the -2 V as the referece point (hence the direction of the arrow, away from the reference point). This gives a potential at that point as -2 + (-8) = -10 (the 8 V across the 60 ohm resistor should be treated as negative in the direction of the arrow as it points in the direction of increasing negative potential).

    So, it is not the resistors that are the wrong way around in your equation, it is the choice of V0.
 
 
 
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