Question

Three identical resistors are connected in series to a battery. If the current of 12 A flows from the battery, how much current flows through any one of the resistors

Answer 1

Answer:

Current that flows through any one of the resistors has a value of 12 amperes.

Explanation:

When a group of resistors are connected in series, the same current flows through each resistor. According to the Ohm's Law, the circuit can be represented as follows:

$V_{batt} = i\cdot (R_{1}+R_{2}+R_{3})$

$i = \frac{V_{batt}}{R_{1}+R_{2}+R_{3}}$

Where:

$V_{batt}$ - Voltage of the battery, measured in volts.

$R_{1}$, $R_{2}$, $R_{3}$ - Electric resistance of the first, second and third resistors, measured in ohms.

$i$ - Current, measured in amperes.

If $R_{1} = R_{2} = R_{3} = R$, then:

$i = \frac{V_{batt}}{3\cdot R}$

Current that flows through any one of the resistors has a value of 12 amperes.

Answer 2

The current flows via any of the resistors should have a value of 12 amperes.

Ohm law:

At the time When a group of resistors are linked in series, so there is a similar current flow via each resistor.

Here the circuit should be

vbatt = i.(R1 + R2+ R3)

i = Vbatt/R1 + R2 + R3

here

Vbatt means the voltage of the battery

R1,R2, and R3 means the resistance of the first, second and third resistors

I means the current

So, in the case when

R1 = R2 = R3 = R

So,

i = Vbatt/3.R

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