Given the circuit diagram below, (a) Find the current, I, the runs through each

Question

Given the circuit diagram below,

(a) Find the current, I, the runs through each of the above resistors.

(b) Assume that the 10 ? resistor just to the right of the 24 V battery is actually an equivalent

resistance for three resistors in parallel. The parallel resistors have resistances 25 ?, 20 ?,

and 100 ?. Given this information, what is the current that flows through each one of these

resistors?

(c) What is the power dissipated in each of the parallel resistors in part (b)?

d) Using your result from part (c), if it takes 333.55 J of heat to melt 1.0 g of ice at 0

Explanation / Answer

Part A)

From Kirchoff's Rules, we will start with the identification of currents

I1 at the top going left

I2 in the middle moving right

I3 at the bottom going left

The two loops, top and bottom, assign clockwise.

Then we have the junction equation

I1 + 13 = 12

The top loop equation...

24 + 18 = -10I1- 10I2 - 15I2

42 = -10I1 - 25I2

The bottom loop equation

6 + 12 -18 = 10I2 + 15I2 + 20I3 + 20I3

0 = 25I2 + 40I3

From this equation -25I2 = 40I3

That means I2 = -1.6I3

From the upper loop equation...

42 = -10I1 - 25I2

42 = -10I1 + 40I3

42 - 40I3 = -10I1

I1 = -4.2 + 4I3

Now from the junction equation...

I1 + 13 = 12

-4.2 + 4I3 + I3 = -1.6I3

-4.2 = -6.6I3

I3 = .636 Amps

I2 = -1.6I3 = -1.02 Amps (The negative just means direction, so I will eliminate it for some other calculations)

I1 = -4.2 + 4(.636) = -1.65 Amps (The negative just means direction, so I will eliminate it for some other calculations)

Thus for the top 10 Ohm Resistor, the current is 1.65 Amps to the right

Through the middle 10 and 15 Ohms resistors, the current is 1.02 Amps to the left

Through the bottom two 20 Ohm resistors, the current is .636 Amps to the left

Part B)

V = IR = (1.65)(10) = 16.5V

Then for the individual resistors.

I = V/R

I = 16.5/20 = .825 Amps through the 20 Ohm Resistor

I = 16.5/25 = .66 Amps through the 25 Ohms Resistor

I = 16.5/100 = .165 Amps through the 100 Ohm Resistor

Part C)

P = IV = (.825)(16.5) = 13.6 W through the 20 Ohm Resistor

P = IV = (.66)(16.5) = 10.9 W through the 25 Ohm Resistor

P = IV = (.165)(16.5) = 2.72 W through the 100 Ohms Resistor

Part D)

P = Energy/time

2.72 = 333.55/t

t = 122.5 sec (That is 2 minutes 2.5 sec)

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