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Two resistors, R1 = 6 Ω and R2 = 12 Ω, are connected in parallel to a 24 V battery. The circuit operates for 5 minutes.

(a) Calculate the total heat generated in both resistors.

(b) If each resistor has a power rating of 100 W, determine whether it is safe to use these resistors in the circuit.

Current Electricity

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Answer

Given,

  • R1 = 6 Ω
  • R2 = 12 Ω
  • Battery voltage (V\text V) = 24 V
  • Operating time of the circuit (t\text t) = 5 minutes = 5 x 60 = 300 seconds

(a) Since the resistors are connected in parallel, the voltage across each resistor is the same as the battery voltage, V\text V = 24 V.

On using Ohm’s Law (I=VR)\left (\text I = \dfrac{\text V}{\text R}\right),

Then, current through R1 is given by,

I1=VR1=246I1=4 A\text I1 = \dfrac{\text V}{\text R1} \\[1em] = \dfrac{24}{6} \\[1em] \Rightarrow \text I_1 = 4\ \text A

And, current through R2 is given by,

I2=VR2=2412I2=2 A\text I2 = \dfrac{\text V}{\text R2} \\[1em] = \dfrac{24}{12} \\[1em] \Rightarrow \text I_2 = 2\ \text A

On using Joule’s Law of Heating (H=I2Rt\text H = \text I^2 \text {Rt}),

Now, heat generated in R1 is given by,

H1=I12×R1×t=42×6×300=16×6×300=16×1800=28800 J\text H1 = \text I1^2 \times \text R_1 \times \text t \\[1em] = 4^2 \times 6 \times 300 \\[1em] = 16 \times 6 \times 300 \\[1em] = 16 \times 1800 \\[1em] = 28800\ \text J

Similarly,

Heat generated in R2 is given by,

H2=I22×R2×t=22×12×300=4×12×300=4×3600=14400 J\text H2 = \text I2^2 \times \text R_2 \times \text t \\[1em] = 2^2 \times 12 \times 300 \\[1em] = 4 \times 12 \times 300 \\[1em] = 4 \times 3600 \\[1em] = 14400\ \text J

Total heat generated = H1+H2\text H1 + \text H2 = 28800 + 14400 = 43200 J

So, the total heat generated in both resistor is 43200 J.

(b) Given,

Power rating of each resistor = 100 W

The power dissipated by each resistor can be calculated using P=V×I\text P = \text V \times \text I

Then,

Power dissipated by R1 is given by,

P1=V×I1=24×4=96 W\text P1 = \text V \times \text I1 \\[1em] = 24 \times 4 \\[1em] = 96\ \text W

Similarly,

Power dissipated by R1 is given by,

P2=V×I2=24×2=48 W\text P2 = \text V \times \text I2 \\[1em] = 24 \times 2 \\[1em] = 48\ \text W

As, R1 is operating at 96 W and R2 at 48 W, which is within the 100 W limit.

Hence, it is safe to use both R1 and R2 in the circuit.

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