1.

Three resistors of 5 Omega, 10 Omega and 15 Omega are connected in series and the combination is connected to the battery of 30 V. Ammeter and voltmeter are connected in the circuit. Draw a circuit diagram to connect all the devices in proper correctorder. What is the current flowing and potential difference across 10 Omega resistance ?

Answer»

Solution :
`R= R_(1)+R_(2)+R_(3)=30 Omega`
`I= V//R= 30//30 =1 A`
V across `10 Omega== IR_(2)=1xx10=10 V`


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