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Question

Three resistors of 6.0Ω,2.0Ω and 4.0Ω are joined to an ammeter A and a cell of e.m.f. 6.0 V as shown in figure. The effective resistance of the circuit is :
178064_df0963fe642c43d9abb6293662c16556.png

A
6.0Ω
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B
2.0Ω
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C
3.0Ω
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D
12.0Ω
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Solution

The correct option is C 3.0Ω
Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. In a series circuit, every device must function for the circuit to be complete. One bulb burning out in a series circuit breaks the circuit. A circuit composed solely of components connected in series is known as a series circuit.The total resistance of resistors in series is equal to the sum of their individual resistances. That is, Rtotal=R1+R2+R3.
Components connected in parallel are connected so the same voltage is applied to each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component. In parallel circuits, each light has its own circuit, so all but one light could be burned out, and the last one will still function. A circuit composed solely of components connected in parallel is known as a parallel circuit.To find the total resistance of all components, the reciprocals of the resistances of each component is added and the reciprocal of the sum is taken. Total resistance will always be less than the value of the smallest resistance That is, 1Rtotal=1R1+1R2+1R3thatis,R=Rtotal1.
In this case, the resistance R2 and R3 are connected in series and the combined resistance is connected in parallel with the resistance R1.
Therefore, the combined resistance of R2 and R3 is given as 4+2=6Ω. The total resistance in the circuit is given as
16+16=10.33 = 3 Ω.

Hence, the effective resistance of the circuit is 3 Ohms.

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