Question

A battery with an emf of 12 V connects in parallel to a 50 Ω and 25 Ω resistor. The current through the battery is 0.70 A. What is the internal resistance of the battery?

Answer 1

r = 0.47 Ω

Step-by-step explanation:

given,

voltage of the battery = 12 V

resistors are connected in parallel.

R₁ = 50 Ω and R₂ = 25 Ω

current,I = 0.70 A

Internal resistance = ?

equivalent resistance

`R_(eq)= (R_1R_2)/(R_1+R_2)`

`R_(eq)= (50* 25)/(50+25)`

`R_(eq)=16.67\ \Omega`

now,

`\varepsilon = I (R_(eq)+r)`

r is the internal resistance

`12 = 0.7* (16.67+r)`

16.67 + r = 17.14

r = 0.47 Ω

Hence, internal resistance of the battery is equal to r = 0.47 Ω