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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