1.The circuit shown in the figure below is connected for 2.50 min. (Assume R 1 =

Question

1.The circuit shown in the figure below is connected for 2.50 min. (Assume R1 = 9.00 , R2 = 1.20 , and V = 10.0 V.)

(a) Determine the current in each branch of the circuit.

(b) Find the energy delivered by each battery.

(c) Find the energy delivered to each resistor.

(d) Identify the type of energy storage transformation that occurs in the operation of the circuit.
14

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(e) Find the total amount of energy transformed into internal energy in the resistors.
15 kJ

branch magnitude (A) direction left branch 1
Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. 2 ---Select--- up down middle branch 3 4 ---Select--- up down right branch 5 6 ---Select--- up down

Explanation / Answer

lets use mesh method to solve the problem

let the current in left mesh be i1 , in clockwise direction and current in right mesh be i2 , in anti clockwise direction

writing KVL in mesh 1:

i1*(R1+5+1)+i2*(5+1)+4=0

15*i1+6*i2=-4....(1)

writing KVL in mesh 2:

i2*(R2+3+5+1)+i1*(5+1)=V-4

6*i1+10.2*i2=6....(2)

solving equations 1 and 2, we get:

i1=-0.656 A

i2=0.974 A

so answers are as below:

a)

left branch : current=i1=0.656 A, down

middle branch=current=i1+i2=0.318 A, down

right branch: current=i2=0.974 A, up

b) energy delivered=voltage*current*time

so for 4 volts battery, energy=4*0.318*2.5*60=190.8 J

as the current is entering into the +ve terminal of 4 volts battery, energy delivered will be negative i.e.. the battery consumes energy.

so energy delivered by 4 volt batery=-190.8 J

for 10 volts battery, current leaves at +ve terminal. hence energy is delivered by the battery.

energy value=10*0.974*2.5*60=1461 J=1.461 kJ

c)energy delivered to each resistors can be found by the formula=energy=current ^2*resistance*time

using the above formula, the following values are obtained for the energy delivered to each resistor:

9 ohm resistor: i1^2*9*2.5*60=581.68 J
5 ohm resistor:(i1+i2)^2*5*2.5*60= 75.189 J
3 ohm resistor:i2^2*2.5*60=142.41 J
1 ohm resistor:(i1+i2)^2*2.5*60=15.164 J
1.2 ohm resistor:i2^2*1.2*2.5*60=170.89 J

d)chemical energy in battery is converted to eelctrical energy which is being dissipated in resistors as heat energy.

e)amount of energy transformed into internal energy=total energy delivered to resistors=581.68+75.189+142.41+15.164+170.89=985.33 J

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