The circuit diagram shown alongside includes a 6 V battery, an ammeter A, a fixed resistor R1, of 2 Ω and resistance wire R2, connected between the terminals A and B. The resistance of the battery and ammeter may be neglected. Calculate the ammeter readings when the wire R2, is of :
(a) 0.20 m length and of resistance 4 Ω.
(b) 0.40 m length and of the same thickness and material as in case (i).
(c) 0.20 m length and having an area of cross section double than that in case (i).
(a) Given,
Battery Voltage (V) = 6 V
Fixed Resistance (R1) = 2 Ω
Resistance of wire (R2) = 4 Ω
R1 and R2 are connected in series.
∴ Total resistance of the circuit (Rs) = R1 + R2 = 2 + 4 = 6 Ω
By Ohm's law,
(b) On doubling the length (i.e., 2 x 0.20 = 0.40 m), the resistance of wire (R2) is doubled, i.e., it becomes 8 Ω.
Total resistance of the circuit (Rs) = R1 + R2 = 2 + 8 = 10 Ω
By Ohm's law,
(c) On doubling the area of cross section, the resistance of wire R2, is reduced to half, i.e., it becomes Ω = 2 Ω.
Total resistance of the circuit (Rs) = R1 + R2 = 2 + 2 = 4 Ω
By Ohm's law,