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Assuming the these resistors are connected in series, the voltage drop across each resists can be calculated simply using V = IR. In a series circuit, the same current must flow through each resistor. So, 2*i + 2*i + 2*i = 6 volts. or i * 6 = 6, or i = 1. So, the voltage drop across all resistors i*r is 2 volts.
Wiki User
∙ 9y ago
Wiki User
∙ 14y agoTotal effective resistance of resistors in series = sum of individual resistances.
(6 + 12 + 24) = 42 ohms.
Voltage in a series circuit divides in proportion to the individual resistances.
9 / (applied voltage) = 12/42
12 x (applied voltage) = 9 x 42
applied voltage = (9 x 42) / 12
= 31.5 volts
Wiki User
∙ 14y agoV=I*R
The question you are asking is flawed in that you only give the resistor values... The voltage drop across these resitors is unknown without knowing either the voltage applied to them, wich would also equal the voltage drop, or the current flowing through them. Using the ohms law equation above all this can be calculated.
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There is insufficient information in the question to answer it. You need to provide either the voltage across the resistor, or the power dissipated by the resistor. please restate the question.
Do you need to apply a load to see a voltage drop across a resistor?
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