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

The internal resistance of the battery is 1.0 Ω.

Step-by-step 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

`(1)/(R)=(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))`

Here, R₁ = R₂ = R₃

`(1)/(R)=(1)/(6)+(1)/(6)+(1)/(6)`

`(1)/(R)=(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)=(V)/(I)`

`R+r=(V)/(I)`

Where, R = resistance

r = internal resistance

V = voltage

I = current

Put the value into the formula

`2+r=(12.0)/(4.0)`

`r=3.0-2`

`r=1.0\ \Omega`

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