In a series electrical circuit, the resistances of the conductors are 2 ohms and 4 ohms. The voltmeter shows a voltage of 12 V. What is the voltage on the first conductor if the second is 8 V?
When the conductors are connected in series, their total resistance is equal to their sum.
Rtot = R + R1
With a series connection, the current flowing through the elements is the same:
I = I1 = I2
The voltage at the ends is equal to the sum of the stresses on the elements:
U = U1 + U2
Let’s find the strength of the current that flows in the second resistance:
According to Ohm’s law, the amount of current is directly proportional to the applied voltage and inversely proportional to its resistance:
I = U / R
I2 = U2 / R2 = 8/4 = 2 A
U1 = I1 * R1 = 2 * 2 = 4V
U = U1 + U2
12 = 8 + 4
It all fits together.
Answer: the voltage on the first element is 4 V.
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