A battery with internal resistance amperes, what is the internal resistance of t

Question

A battery with internal resistance 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 12 ? resistor. How much current flows through this resistor? (e) Hov much power is dissipated in the 12 2 resistor? f The leads to a voltmeter are placed at the two ends of the battery of this circuit containing the 12 ? resistor. What does the meter read? 18

Explanation / Answer

Given , E = 9V , I = 18 A
Part A) Here , External resistance (R) = 0
Using , I = E / ( R + r)
18 = 9 / ( 0 + r )

r = 9/18 = 0.5 ohms
Internal resistance, r = 0.5 ohms

b) Power(P) = I^2 r = 18^2 *0.5 = 162 W

c) Energy(E) = Power * time = 162 W*1 s = 162J
(d) Current, I' = E / ( R + r ) = 9 / ( 12 + 0.5 ) = 0.72 A
(e) Power, P = I'^2 * ( R ) = 0.72^2 *12 = 6.221 W
(f) V = E - i'*r = 9 - [ 0.72* (0.5) ] = 8.64 V

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