A battery of EMF 15 volt and internal resistance 2 ohm is connected to two resistors of 4 ohm and 6 ohm joined (a) in series, (b)in parallel. Find in each case the electrical energy spent per minute in 6 ohm resistor.
Answers
Answer:
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