Question

(1) A battery of e.m.f. 15V and internal resistance 2 ohm is connected to two resistors of resistance 4 ohm and 6 ohm joined in series. Find the electrical energy spent per minute in 6 ohm resistor?
(2) Water in an electric kettle connected to a 220V supply took 5 minutes to reach its boiling point. How long would it have taken if the supply voltage had fallen to 200V?
(3) You have three resistors of values 2 ohm, 3 ohm and 5 ohm. How will you join them so that the total resitance is less than 1 ohm? Draw a diagram and find the total resistance.

Answer 1

EMF = 15 V
Total resistance in the circuit = 2 + 4 + 6 = 12 Ω
Current = I = 15/12 = 5/4 = 1.25 A

Energy in 1 min = Power * time duration
   = I² R * T
   = 5²/4² Α² * 6 Ω * 60 sec
   = 562.50 Joules
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2)
Heat energy supplied to water = power * duration
   Q =  V² / R * T
   So  T =  Q * R / V²
 
 since Q, T are same, 
   T2 / T1 =  V1² / V2²
   T2 / 5 min  = 220² / 200²     
   T2 = 1.21 *5 =  6.05 minutes    6 minutes,  3 sec
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3)
   R1 = 2 Ohm     R2 = 3 Ohm        R3 = 5 Ohm
total resistance R should be  < 1 Ohm.
 
If any resistance is connected in series, then the total resistance will be more than that resistance.  So effective resistance will be more than 2 Ohm or 3 Ohm or 5 ohm..

Hence,  Connect all of them in parallel. 
   Effective resistance = R
       1/R = 1/2 + 1/3 + 1/5  =  31/30
     R = 30/31