Two identical batteries, each with an emf of 18 V and an internal resistance of 1 :, are wired in parallel by connecting their positive terminals together and connecting their negative terminals together. The combination is then wired across a 4-: resistor. The current in the 4-: resistor is:

When the batteries are connected in parallel, the voltage is the same as the voltage of each battery, but the internal resistance is less, in general we can assume that the internal resistance is also in parallel.

1 / [tex]r_{eq}[/tex] = 1 / r₁ + 1 / r₂ = 2 / r = 2/1

r_{eq} = 0.5 ohm

The current in the resistor is

V = I R + I r_{eq}

I = V / R + r_{eq}

I = 18 / (4 +0.5)

I = 4 A

priyankatutor priyankatutor · Answer 2 · 2021-12-24T10:26:05+01:00

The current in the resistor is 4 A, when batteries are connected in parallel.

If batteries are connected in parallel, the voltage is the same in each battery, but the internal resistance is less.

Assume the internal resistance,

[tex]\bold {\dfrac 1 {R_e_q} = \dfrac 1 {R_1 } + \dfrac 1 {R_2} }}\\\\\bold {\dfrac 1 {R_e_q} = \dfrac 21}\\\\\bold { R_{eq} = 0.5\ ohm }}[/tex]Since R = 1

The current in the resistor,

[tex]\bold { V = I R + I R_{eq}}\\\\\bold { I = \dfrac V R + R_{eq}}\\\\\bold { I = 18 { (4 +0.5)}}\\\\ \bold { I = 4\ A}[/tex]

Therefore, the current in the resistor is 4 A, when batteries are connected in parallel.

To know more about internal resistance,

https://brainly.com/question/19690382