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Q) Two resistors of resistances 2 Ω and 4 Ω are connected in

  1. a) series
  2. b) parallel

with a battery of given potential difference. Compute the ratio of total quantity of heat produced in the combination in the two cases if the total voltage and time are kept the same for both.

(CBSE Sample Paper – 2024-25)

Ans: 

STEP BY STEP SOLUTION

Let’s consider the Battery Voltage = V and time is T seconds

Step 1: Case A – Resistances 2𝛺 and 4𝛺 are connected in series:

Equivalent resistance of the series combination = 2 + 4 = 6 Ω

Power consumed b the series combination, PS = \frac{V^2}{R} = \frac{V^2}{6}

Heat produced in T seconds, HS = Ps x T = \frac{V^2}{6} x T = \frac{V^2 T}{6}

Step 2: Case B – Resistances 2𝛺 and 4𝛺 are connected in Parallel:

Equivalent resistance of the parallel combination = \frac{2 \times 4}{2 + 4}

= \frac{8}{6} = \frac{4}{3} Ω

Power consumed b the parallel combination, PP = \frac{V^2}{R}

= \frac{V^2}{\frac{4}{3}} = \frac{3 V^2}{4}

Heat produced in T seconds, HP = PP x T = \frac{3 V^2}{4} x T = \frac{3 V^2 T}{4}

Step 3: Ratio of heats:

Heat during Series connection : Heat during parallel connection = HS : HP

∴ \frac{H_S}{H_P} = \frac{\frac{V^2 T}{6}}{\frac{3 V^2 T}{4}}

=  \frac{4}{6 \times 3} = \frac{2}{9}

∴  HS : HP :: 2 : 9

Therefore, the ratio of total quantity of heat produced in the combination in the two cases is 2: 9

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