Question

Three resistors, 2.62, 3.86, and 4.28Ω, are connected in series across a 15.3-V battery. Find the power delivered to each resistor.?

Answer 1

solution:-

given data

resistance(R₁)=$2.62 \Omega$

resistance(R₂)=3.86$\Omega$

resistance(R₃)=4.28$\Omega$

voltage(V)=15.3V

all the resister are in series so equivalent resistance can obtain as

$R=R_1+R_2+R_3$

by putting the value

$R=2.62+3.86+4.28=10.76\Omega$

all the resister are in series so current will be same in all resistance

current(I)=$\frac{V}{R}$

by putting value of V and R

$I=\frac{15.3}{10.76}=1.42A$

power in any resister can be obtain as

$P=I^2R$

power in resistance R₁

$P_1=1.42\times1.42\times2.62=5.28W$

power in resistance R₂

$P_2=1.42\times1.42\times3.86=7.76W$

power in resistance R₃

$P_3=1.42\times1.42\times4.28=8.60W$

so power in R₁ ,R₂ ,and R₃ are 5.28W , 7.76W and 8.60W respectively.