In the figure below, R1 = 6.70 and R2 = R3 = R4 = 25.0 . The battery has an emf
ID: 1427172 • Letter: I
Question
In the figure below, R1 = 6.70 and R2 = R3 = R4 = 25.0 . The battery has an emf of = 120 V.
a.) How much energy is dissipated by all four resistors in one minute?
(Hint: First, find the total power dissipated.)
b.) What are the current flowing through and voltage drop across R1?
I1 =
V1 =
Also, find the common voltage and current for each of the other resistors. Comment on how they would compare (Choose happen to be equal or will always be equal for I and V) if these resistors did NOT happen to have the same resistance.
I2 = I3 = I4 =
V2 = V3 = V4 =
Explanation / Answer
let,
battery potential V=120v
R1=6.7 ohms
R2=R3=R4=25 ohms
a)
here,
R2, R3, R4 are in parallel combination,
1/R234=1/R2 + 1/R3 + 1/R4
1/R234=1/25 + 1/25 + 1/25
====> R234=8.3 ohms
and
R1, R234 are in series combination,
R1234=R1+R234
=6.7+8.3
=15 ohms,
now,
current i1234=v/R1234
=120/15
=8 A
power p=i_1234^2*R
=8^2*15
=960 w
enegry U=P*t
=960*60
U=57600 J
U=57.6 kJ
b)
current through R234, R1 is,
i1=i234=8A
now,
v1=i1*R1
v1=8*6.7
v1=53.6 v
potentail across R1 is v1=53.6 v
current through R1 is i1=8 A
and
v234=i234*R234
=8*8.3
=66.4 v
but,
v2=v3=v4=v234=66.4 v
now,
i2=V2/R2=66.4/25 =2.66 A
i3=V3/R3=66.4/25 =2.66 A
i4=V4/R4=66.4/25 =2.66 A