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Compare the power used in the 2 Ω resistor in each of the following circuits:
(i) A 6 V battery in series with 1 Ω and 2 Ω resistors
(ii) A 4 V battery in parallel with 12 Ω and 2 Ω resistors
(i) Potential difference V1 = 6V ; Resistance R1= 1Ω and R2= 2Ω
As per ohm's law
V=IR
P=I2R
Equivalent Resistance= R1+R2 (connected in series)
R = 1 + 2
R= 3Ω
As per ohm's law
V =IR
I = V1 /R
I = 6/3
I = 2A
Since there is no division in the series circuit, the current will pass through every component. 2Ω resistor has a current of 2A will flow through it as a result.
Since the current remains constant
P1 = I2R
P1 = 22 X 2
P1 = 8 W
(ii) Potential difference V2 = 4V ; Resistance R3= 1Ω and R4= 2Ω
Potential difference V2 = 4V
Since it is a parallel circuit, the voltage will remain constant. The voltage across 2Ω resistor will be 4V.
P2 = (V2)2/R
P2 = (4)2/2
P2 = 8 W
Power used by both circuits is 8 W.
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