In the circuit below, the ammeter has an internal resistance of 100.0 ohms and t

Question

In the circuit below, the ammeter has an internal resistance of 100.0 ohms and the voltmeter has a sensititvity of 30.00 k Ohm per Volt and is being used on the 10 Volt scale. The battery is a "real battery", which has an internal resistance of 2.0 ohms and an unknown EMF. The voltmeter reads 3.998 Volts, and the ammeter reads 0.01999 mA. The current in the unknown resistance R_x equals 0.01999 mA 0.1333 mA 0.01333 mA 0.00666 mA 0.009995 mA The unknown resistance R_x is equal to 100 k Ohm 200 k Ohm 300 k Ohm 500 k Ohm 600 k Ohm. The potential difference across the terminals of the battery from point "a" to point "b" is equal to 5.997 V 8.997 V 9.000 V. 3.998 V 4.998 V.

Explanation / Answer

a)

Same current flows through the resistance which are connected in series.So same flows through Resistance Rx.

I=0.01999 mA

b)

Unknown resistance

Rx=V/I =3.998/(0.01999*10-3)

Rx = 200*103 ohms= 200 Kohms

c)

equivalent resistance

Req =2+250000+200000 =450002

Potential difference across the battery

Vab=IReq =450002*0.01999*10-3

Vab = 8.997 Volts

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