1.

Draw a circuit diagram for a circuit consisting of a battery of five cells of 2 volts each, a 5 Omegaresistor, a 10 Omega resistor and a 15 Omega resistor, an ammeter and a plug key, all connected in aeries. Also connect a voltmeter to record tht: potential difference across the 15 Omega resistor and calculate : (i) the electric current passing through the above circuit and (ii) potential difference across 5 Omega resistor when the key is closed.

Answer»

SOLUTION :(i)
10 V battery Rest COMPONENTS
Equivalent resistance =`R_(1)+R_(2)+R_(3)`
`=5+10+15`
`=30 Omega`
Current in the CIRCUIT `I= (V)/(R)`
`I = (10V)/(30Omega) = (1)/(3) A ` or 0.33 A
(ii) Potential difference across 5 `Omega` resistor V= IR
`=(1)/(3)A xx5 Omega`
1.67 V


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