You short-circuit a 22 volt battery by connecting a short wire from one end of the battery to the other end. If the current in the short circuit is measured to be 18 amperes, what is the internal resistance of the battery? What is the power generated by the battery? How much energy is dissipated in the internal resistance every second? (Remember that one watt is one joule per second.) This same battery is now connected to a 6 OHM resistor. How much current flows through this resistor? How much power is dissipated in the 6 OHM resistor? The leads to a voltmeter are placed at the two ends of the battery of this circuit containing the 6 OHM resistor. What does the meter read?
Voltage V = 22 volt
Current i = 18 A
Internal resistance r = V/ i
= 22 / 18
= 1.222 ohm
(b).Power generated P = i 2 r
= 18 2 (1.222)
= 396 watt
(c).Energy is dissipated in the internal resistance every second E = i 2 r t
Where t = 1 s
So, E = 18 2 (1.2222) (1)
= 396 J
(d).Resistance R = 6 ohm
Current i \' = V/ ( r +R )
= 22 /( 1.2222+6)
= 3.046 A
(e).Power dissipated in the 6 ohm resistor P \' = i \' 2 R
= 3.046 2 ( 6)
= 55.67 watt
(f). Required reading V \' = V - i\'R
= 22 -(3.046x6)
= 3.724 volt
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