Answer :

Given:


Emf,


External resistor,


Internal resistance,


Resistance of Ammeter,


Resistance of Voltmeter,


(a)


Concepts/Formula used:
Resistors in Series:


Resistors in parallel:


Ohm’s Law:


Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:



We are trying to find the equivalent resistance of the circuit first.



We will treat the measuring devices as ordinary resistors. Now, voltmeter and R= 50Ω are in parallel. The equivalent resistance is given by:




We can redraw the circuit as follows:



Now, we can see that all the devices are in series.


Hence,





Now, the current that passes through Req also passes through the three series components : internal resistor, r’ and ammeter.


Hence,


Using Ohm’s law,




Now, the potential difference across R is the same as potential difference across r’.




Hence, the ammeter reading is 0.1A and the voltmeter reading is 4V.


(b)


Concepts/Formula used:
Resistors in Series:


Resistors in parallel:


Ohm’s Law:


Potential Difference (V) across a resistor of resistance R when current I passes through it is given by Ohm’s law:



Let us find the equivalent resistance, Req of the circuit.



We will treat the ammeter and the voltmeter as ordinary resistors to find the equivalent resistance.


Note that the ammeter and the external resistor are in series.


Hence,



We can redraw the circuit as follows:



Note that the voltmeter and r’ are in parallel.




The circuit can be redrawn as follows:



Now, the r’’ and the internal resistance are in series.



By Ohm’s law, the current coming out of the battery is



Now, potential difference across r’’ = 41.27 Ω is the same as across r = 52 Ω .


Voltmeter Reading:



Now, the current passing through the ammeter is the one passing through r’.



The ammeter reading is 0.08A and the voltmeter reading is about 4.3 V.


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