Answer to Question #228176 in Electric Circuits for Ajiwaq

Question

A battery is connected in series with a variable resistor R and an ammeter in order to form a circuit. The resistor of the ammeter may be neglected. When R is given the value 1.1ohm the current in the circuit is 4.0amp but when the R is given value 1.6ohm the current in the circuit is 3.0amp. Find the internal resistance of the battery

Answer 1

Gives

R=1.1 ohm

I=4.0 ohm

$V=iR\\V=4\times1.1=4.4Amp\\\\E=3.0\times1.6=4.8ohm \\V=E-ir$

Put value

$4.4=4.8-4.0\times r$

$4.0r=4.8-4.4=0.4$

$R=\frac{0.4}{4}=0.1\Omega$

Internal resistance(r)=0.1 ohm