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Draw a schematic diagram of a circuit consisting of a battery of 3 cells of 2V each,a combination of three resistors of 10 Omega, 20 Omega and 30 Omega connected in a parallel, a plug key and an ammeter, all connected in series. Use the circuit to find the value of the following :(a) current through each resistor (b) total current in the circuit. (c )total effective resistance of the circuit.

Answer»

Solution :The circuit diagram is drawn hare. HEREB is a battery of 3 cells of 2V each i.e. having a total voltage of 6 V, three resistors `R_(1) = 10 Omega, R_(2) = 20 Omega, R_(3) = 30 Omega` connected in a parallel across points P and Q a plug key K and and ammeter A all joined in series.
(a) Current through resistors, `R_(1), I_(1) = (V)/(R_(1)) = (6V)/(10 Omega) = 0.6 A`.
Current through resistors, `R_(2) , I_(2)(V)/(R_(2)) = (6V)/(20 Omega) = 0.3 A`.
Current through resistors,`R_(3) , I_(3)(V)/(R_(3)) = (6V)/(30 Omega) = 0.2 A`.
(b)Total current in the circuit `I = I_(1) + I_(2) + I_(3) = 0.6 + 0.3 + 0.2 = 1.1A`
(c )total EFFECTIVE resistance of the circuit, `R = (V)/(I) = (6V)/(1.1A) =(60)/(11) Omega = 5.5 Omega`


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