Welcome to the NCERT Solutions for Class 10 Science - Chapter Electricity. This page offers a step-by-step solution to the specific question from Exercise 8, Question 4: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –
(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1
(c) The Joule heating is given by, H = i2Rt
Let, R be the resistance of the two wires.
The equivalent resistance of the series connection is RS = R + R = 2R
If V is the applied potential difference, then it is the voltage across the equivalent resistance.
The heat dissipated in time t is,
The equivalent resistance of the parallel connection is
V is the applied potential difference across this RP.
The heat dissipated in time t is,
So, the ratio of heat produced is,
Note: H 
also H i2 and H t. In this question, t is same for both the circuit. But the current through the equivalent resistance of both the circuit is different. We could have solved the question directly using H R if in case the current was also same. As we know the voltage and resistance of the circuits, we have calculated i in terms of voltage and resistance and used in the equation H = i2Rt to find the ratio.
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Read MoreWelcome to the NCERT Solutions for Class 10 Science - Chapter . This page offers a step-by-step solution to the specific question from Excercise 8 , Question 4: Two conducting wires of the same material and of equal lengths and equal diameters are first connect....
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