In the circuit shown, the 6.0 ohm resistor is consuming energy at a rate of 36 J

Question

In the circuit shown, the 6.0 ohm resistor is consuming energy at a rate of 36 J/s when the current through it flows as shown. Find,

If you can, please show the math for the question. I've been struggling on it. Thank you!

3. In the circuit shown in Fig. shown below, the 6.0 resistor is consuming energy at a rate of 36 J s when the current through it flows as shown. Find, (a) The curent through the ammeter A (b) What are the polarity and voltage ofthe unknown battery, assuming it has negligible internal resistance? 20.0 17 13 20.0 =25 v 20.012 19 3.0 6.0 1.0

Explanation / Answer

I'll replace the 2-parallel connected resitors with the equivalent resistance Re=10 Ohms (Re=R/2).

We also know: the power of 6 Ohms resistor is: P(6)=I2*R=I2*6 ---> I2=P/R=6 --> I~2.45 A.

Write Kirchhoff's laws for the node and 2 loops (I1-through Re, 19 Ohms, Ammeter, 1 Ohm loop; I2-through :13 Ohms, unknown E, 3 Ohms loop); I chose to travel on each loop clockwise.

I=I1+I2 --> I2=I-I1

25-E=17*2.45+(13+3)I2+6*2.45

E=-(13+3)I2+(10+19+1)I1

Solve this system of eqs (replace I2 in the last 2 eqs, add them, reduce the corresponding terms) and get:

I1=-1.352 A

I2=~3.801 A

Replace I1,2 and get E: ~-87.244 V

The signs tell that I1 and E have opposite sense/polarity.

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