Question

A battery of 12.0V powers a circuit in which two resistance R1 = 150Ω e R2 = 300Ω are linked in parallel. Compute the current intensity circulating in the circuit.

Answer 1

Given:

Voltage=12V

R₁=150 $\Omega$

R₂=300$\Omega$

In parallel combination equivalent resistance is given by

${1\over R_eq}={1\over R_1}+{1\over R_2}$

On putting values we gi et,

R_eq= 100$\Omega$

And according to ohms law

$\boxed{V=iR}$

from this we get

$I={12\over100}\\$

$\boxed{=0.12 Amp.}$

Therefore intensity of current in circuit is 0.12 A.