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?
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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