(30 points) A 110.00 (v) batter is connected to a resistor network as shown belo
ID: 2064261 • Letter: #
Question
(30 points) A 110.00 (v) batter is connected to a resistor network as shown below in fig. (i). (8 points) What's the equivalent resistance Rp of the circuit in fig.(ii)? (4) What's the current I1 through the 400.0-ohm resistor? (6) What's the rate of heat dissipation (in Watts) within the 400.0-ohm resistor? (4) What's the current I2 through the 300.0 ohm resistor? (4) What's the rate of heat dissipation (in Watts) within the 300.0-ohm resistor? (4) How much heat energy (in Joules) is produced in the 300.0-ohm resistor in 60 seconds?Explanation / Answer
as both the resistances are in parallel so the equivalent resistance will be
a. Req=400x 300/(400+300)
Req=171.42
b. by ohm's law
Ieq=V/Req
Ieq=110/171.42
Ieq=.641 A
current through 400 ohm resistor we use current division rule
I400=Ieq x (300)/(300+400)
I400= .275A
current through 300 ohm resistor will be
I300=Ieq(400)/(300+400)
I300=.36A
heat dissipation in 400 ohm resistor
P=I2R
=(.275)2 x (400)
= 30.25 Watt
Heat dissipation in 300 ohm resistor
P=I2R
=(.36)^2 x (300)
= 38.88 Watt
energy in 60 sec in 300 ohm resisitor
E= Pxt
= 38.88 x 60
E=2332.8 J
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