the ones in red are the answers, but i need the steps Problem 11: Q4) A resistor

Question

the ones in red are the answers, but i need the steps

Problem 11: Q4) A resistor R is connected to a battery with emf (EF V), and an internal resistance r i Q as shown in the figure. r R-19Q a) When th switchsisopen, what is the terminal voltage Lethereadingofthe voltmeter VO? The switch Sisdosed now. (ie thereadingof thevoltmeterV and b) What is the (ie, thereading of the Ammeter (A)? c) The internal resistance (r) of the battery increases after the battery is used and as a result the current decreases. When the current through the resistor dropped to the half of the value you calculated in part (b) above, find: the em and the internal resistance Cr of the battery. d) What is the rate at which energy is provided by the battery (power)? a 1V b) I .05A V-0.95V c) 212 d) 025W

Explanation / Answer

part 1)

When circuit is open then there will no current flow through the 19 ohm resister. So all emf that is 1V will be drop through thermal voltage.

So ans will same as emf i.e. 1V.

part 2)

Now switch is closed so current will flow through resister 19 ohm.

total equivelent resistance (here series connection ) of the circuit is given by

1ohm+19ohm=20ohms

current =1V/20ohm=0.05ampere # ameter will show 0.05A reading

now calculate the potential drop on the internal resistance that is 1 ohm, is given by

0.05A * 1 ohm=0.05V

so only 1V-0.05V=0.95V will be shown by voltameter.

part 3)

emf would be same

final current =0.05A/2=0.025A

0.025*(19+x)=1V

=>x=21ohm #this is the final internal resistance

part 4)

power=current * voltage

power=0.025A*1V=0.025W

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