How much energy is dissipated as heat during a two-minute time interval by a 1 .

Question

How much energy is dissipated as heat during a two-minute time interval by a 1 .5-k ohm resistor which has a constant 20-V potential difference across its leads? a) 58 J b) 46 J c) 32 J d) 72 J e) 17 J When a voltage VI is placed across a series combination of an unknown resistor R and a 65-ohm resistor, the power delivered to the resistors is Pl. When the 65-ohm resistor is replaced by a 20-ohm resistor (and the voltage is kept the same) the power delivered to the resistors doubles to a value of 2PI. What is the unknown resistance? a) 75 ohm b) 55 ohm c) 40 ohm d) 25 ohm e) 10 ohm A resistor of unknown resistance and a 15-ohm are connected to a 20-V emf. A 2.0 A current is observed to he passing through the emf. What is the value of the unknown resistance? [Note that it is intentionally not specified that the resistor combination is series or parallel; though it's not part of the answer, you also need to figure that out.] 75 ohm 30 ohm 7.5 ohm 12 ohm 6.0 ohm

Explanation / Answer

12) Energy = power x time

power = p = VI = V2/R =0.266 W

Energy = 0.266 x 2 x 60 = 32 J

13) resistor are connected in series

hence Power P1 = V2/(R+65)

when the 65 ohms is replaced by 20 ohms

Power P2 = V2/(R+20)

but given P2 = 2P1

1/(R+20) = 2/(R+65)

R + 65 = 2R + 40

R = 25 Ohms

14) Since the current is 2 A

Lets calculate the drop across the 15 ohm resistor

V' = 15 x 2 = 30V Since this is higher than the applied voltage, 2 A current is not passing through 15 ohm resistor.Hence the resistors are connected parallely

Now effective resistance R'= 15R/(R+15) for parallel connection

Now voltage = iR'

20 = 30R/(R+15)

20R +300 = 30R

R = 30 ohms

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