Two identical resistors each of resistance 2 Ohm are connected in turn (1) in series (2) in parallel to a battery of 12 V. Calculate the ratio of power consumed in two cases.
Given; The two resistors R1=R2=2ΩR1=R2=2Ω . The potential of the battery, V=20VV=20V . (i) Considering the first case in which the two resistors are connected in series, the resultant resistance of the connection is given as, R=R1+R2R=R1+R2 On substituting the values in the above expression, we get, R=(2+2)ΩR=4ΩR=(2+2)ΩR=4Ω The current that will be flowing through each resistor will be, I=VRI=124I=3AI=VRI=124I=3A So, the power consumed in this case is, P1=V⋅IP1=12×3P1=36WP1=V⋅IP1=12×3P1=36W Therefore, the power consumed when the resistors are connected in series is 36 W. (ii) Considering the second case in which the two resistors are connected in parallel, the resultant resistance of the connection is given as, 1R=1R1+1R21R=1R1+1R2 On substituting the values in the above expression, we get, 1R=(12+12)Ω1R=1Ω1R=(12+12)Ω1R=1Ω The current that will be flowing through each resistor is, I=VRI=121I=12AI=VRI=121I=12A So, the power consumed in this case is, P2=V⋅IP2=12×1P2=12W Therefore, the power consumed when the resistors are connected in series is 12 W. Hence, the ratio of the powers is calculated as, P1P2=3612P1P2=31P1P2=3:1P1P2=3612P1P2=31P1P2=3:1 Therefore, the ratio of the power consumed is 3:1 .
