Question

The terminal voltage of a real battery that delivers 16.2 W of power to a load resistor is 13.3 V, and its emf is 16.2 V. What is the resistance of the following?

(a) the load resistor in this circuit
(b) the internal resistor of the battery

Answer 1

`r=2.3808\ \Omega`

Step-by-step explanation:

Given:

• power of battery,
  `P_b=16.2\ W`

• voltage across the load resistor,
  `V=13.3\ V`

• emf of the battery,
  `emf=16.2\ V`

Now as we know that the electric power is given as:

`P_b=(V^2)/(R)`

here:
`R=` resistance of the load

a)

Now the load resistance:

`R=(V^2)/(P_b)`

`R=(13.3^2)/(16.2)`

`R=10.92\ \Omega`

b)

Using Ohm's law, current in the load:

`I=(V)/(R)`

`I=(13.3)/(10.92)`

`I=1.218\ A`

Now,

`emf-V=r.I`

where:
`r=` internal resistance

`16.2-13.3=r* 1.218`

`r=2.3808\ \Omega`