Question

T.e.m a unei surse de curent este de 12 V iar rezistenta ei interioara de 0,5 ohmi.la ce rezistenta exterioara a circuitului intensitatea va fi egala cu 2 A.

Answer 1

Answer:

$5.5\Omega$

Explanation:

Here we want to find the external resistance of the circuit.

First of all, the terminal voltage of a battery is given by:

$V=\epsilon - Ir$

where

$\epsilon$ is the emf

I is the current

r is the internal resistance

According to Ohm's law, when the battery is connected to the external resistance, the voltage on the battery is equal to the voltage drop on the resistor, which is:

$V=IR$

where

R is the resistance of the external resistor

Combining the two equations,

$\epsilon - Ir = IR$

Here we have:

$\epsilon=12 V$ is the emf of this battery

$r=0.5 \Omega$ is the internal resistance

I = 2 A is the current in the circuit

Solving for R, we find:

$R=\frac{\epsilon -Ir}{I}=\frac{12-(2)(0.5)}{2}=5.5\Omega$