questions for face-to-face teaching purposes only. Two light bulbs. one rated 30

Question

questions for face-to-face teaching purposes only. Two light bulbs. one rated 30 W al 120 V and another rated 40 W at 120 V, are arranged in two different circuits. The two bulbs are first connected in parallel lo a 120 V source. Determine the resistance of the bulb rated 30 W and the current in it when it is connected in this circuit. Determine the resistance of the bulb rated 40 W and the current in it when it is connected in this circuit. The bulbs are now connected in series with each other and a 120 V source. Determine the resistance of the bulb rated 30 W and the current in il when it is connected in this circuit. Deiermine the resistance of the bulb rated 40 W and the current in it when il is connected in this circuit. In the spaces below, number the bulbs in each situation described, in order of their brightness. (1 = brightest, 4 = dimmest) 30 W bulb in the parallel circuit 40 W bulb in llie parallel circuit 30 W bulb in the series circuit 40 W bulb in the series circuit Calculate the toial power dissipated by ihe two bulbs in each of the following cases. The parallel circuit The scries circuit

Explanation / Answer

The given bulbs are P1 = 30W at V1 = 120V then resistance of the bulb R1 = V1^2 /P1                                                = (120V)^2 / 30W = 480 and   P2 = 40W at V2 = 120V then the resistance of the bulb is                          R2 = V2^2 /P2                               = (120V)^2 / 40W = 360 (a) When they connect in parallel            the eqivalent resistance R' = R1R2 /R1 + R2                                                    = (480)(360) / 480 + 360                                                    = 205 so then the current I = V/R'                                = 120V/ 205                                = 0.58 A The current through the first resistor R1 is I1 = V1/R1 = 120V/480 = 0.25A then the current through second resistor is R2 is I2 = 0.33 A (b) When they are connected in series        the equivalent resistance                  R' = R1 + R2 = 480 + 360 = 840         then the current I = V/R'                                    = 120V / 840 = 0.14 A There is same current through the two resistors (c) The brightness of the bulb is depends upon the current and the resistance, since time is same for all bulbs         so for 30W H = (0.25)^2 (480) = 30      2       so for 40W    H = (0.33)^2 (360) = 39      1 and in series            for 30W    H = (0.14)^2(480) = 9.4       3            for 40W    H = (0.14)^2 (360) = 7       4 therefore the buld of 40W in parallel have bright (d) The power disspated in parallel ciruit         P1 = 30W and P2 = 39W in series circuit P1 = 9.4W and P2 = 7W                                                     

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