Question

A simple circuit has a 12.0V12.0V battery and three resistors of 6.0Ω6.0Ω connected in parallel. If the current through the battery is 4.0A4.0A , what is the internal resistance of the battery?

Answer 1

Answer:

The internal resistance of the battery is 1.0 Ω.

Explanation:

Given that,

Voltage = 12,0 V

Resistance of each resistor = 6.0 Ω

Current = 4.0 A

When three resistors are connected in parallel.

We need to calculate the resistance

Using formula of parallel

$\dfrac{1}{R}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$

Here, R₁ = R₂ = R₃

$\dfrac{1}{R}=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}$

$\dfrac{1}{R}=\dfrac{1}{2}$

$R = 2\ \Omega$

The equivalent resistance is

$R_{eq}=R+r$

$R_{eq}=2+r$

We need to calculate the internal resistance

Using ohm's law

$V = IR$

$R_{eq}=\dfrac{V}{I}$

$R+r=\dfrac{V}{I}$

Where, R = resistance

r = internal resistance

V = voltage

I = current

Put the value into the formula

$2+r=\dfrac{12.0}{4.0}$

$r=3.0-2$

$r=1.0\ \Omega$

Hence, The internal resistance of the battery is 1.0 Ω.