In the circuit shown in the figure (Figure 1) . the 6.0 ohm resistor is consuming...

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In the circuit shown in the figure (Figure 1) . the 6.0 ohm resistor is consuming...

In the circuit shown in the figure (Figure 1) . th

In the circuit shown in the figure (Figure 1) . the 6.0 ohm resistor is consuming energy at a rate of 23.0 J/s when the current through it flows as shown. Find the current through the ammeter A. What are the polarity and emf of the battery , assuming it has negligible internal resistance?

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Answer 1

Answer #1

Current through Resistance 6:

P=VI =I²*R

23=I²*6

I = 1.9579 A

Equivalent resistance of two parralel 20 Ω =(1/20+1/20)^-2 =10 Ω

Let current though Ameter is I1.

By kirchoff law;

-6*1.9579-10I1 - 19*I1-I1-3*1.9579+25 = 0

-30*I1-17.6211+25=0

30*I1=7.3789

I1=0.24596 A

the current through the ammeter A is 0.24596 A.

Again by kirchoff law:

Current through E

I2 = 1.9579-0.24596 = 1.71194 A

Polarity is above negative and Lower positive.

By kirchoff law:

-30*1.71194-9*1.9579+25+E=0

E = 43.98 V OR 44 V

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Answer 2

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