A battery with internal resistance (a) You short-circuit a 9 volt battery by con

Question

A battery with internal resistance (a) You short-circuit a 9 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 20 amperes, what is the internal resistance of the battery? (b) What is the power generated by the battery? (c) How much energy is dissipated in the internal resistance every second? (Remember that one watt is one joule per second.) (d) This same battery is now connected to a 6 n resiston. How much current flows through this resistor? (e) How much power is dissipated in the 6 n resistor? (f) The leads to a voltmeter are placed at the two ends of the battery of this circuit containing the 6 n resistor. What does the meter read?

Explanation / Answer

a) R_in = V/I

= 9/20

= 0.45 ohms

b) P = I*V

= 9*20

= 180 Watts

c) Thermal energy dissipated per second = P*time

= 180 J/s *1 s

= 180 J

d) I = V/Rnet

= 9/(6+0.45)

= 1.4 A

e) P(6) = I^2*R

= 1.4^2*6

= 11.7 watts

f) V_terminal = Vo - I*Rin

= 9 - 1.4*0.45

= 8.37 volts

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