Answer to Question #276804 in Electric Circuits for Berry

Question

Two filament lamp A and B takes 0.8Ampere respectively when connected across 110 volt supply.Determine the value of current when they are connected in series

Answer 1

I=0.8Amp

V=110 volt

Lamp A resistance = R₁

Lamp B resistance=R₂

equlalent resistance

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$R_{eq}=\frac{R_1\times R_2}{R_1+R_2}$

Puut value

Now each resistance current flo 0.8amp then both resistance are equal

$R_1=R_2=R$

$R_{eq}=\frac{R\times R}{R+R}=\frac{R}{2}$

Ohm's law

$V=IR_{eq}$

$R_{eq}=\frac{V}{I}$

$R=\frac{2V}{I}$

Put value

$R=\frac{2\times110}{0.8}=275 \Omega$

Now both filament are series combination

$R_{eq}=R_1+R_2$

$R_{eq}=275+275=550\Omega$

Now calculate current

Series combination current floweach bulb is same

$I=\frac{V}{R_{eq}}$

$I=\frac{110}{550}=0.2Amp$

Each bulb current flow

$I_1=I_2=0.2Amp$