The heat generated in 5Ω resistor due to current flowing through it is 10 cal per sec. The heat generated in 4Ω resistor (in cal/sec) is _________

Solution:

Resistance in upper branch is 10 $Ω$ and in lower branch is 5 $Ω$ . Thus, if current in lower branch is i, current in upper branch is (i/2).

$\begin{array}{l} Now H_{4} = (i / 2)^{2} (4) \\ H_{5} = i^{2} (5) \\ \frac{H_{4}}{H_{5}} = \frac{1}{5} \\ or H_{4} = \frac{H_{5}}{5} = \frac{10}{5} = 2 cal / \sec \end{array}$

Related content

Test your English Vocabulary

CUET Exam Dates 2024 – Application Form, Fees, Eligibility

CBSE Class 12 IP Answer Key 2024,Informatics Practices Paper Solution For SET 1, 2, 3, 4

CUET UG Cut Off 2024, Category, Universities and Colleges Wise Expected Cut Off

Modal Verbs

Helping Verbs

Letter To Your Friend About Your School Trip

Action Verbs

CUET 2024 – List of Colleges and Participating Universities Accepting CUET Exam Score

SRMJEEE Online Test Series – Practice Papers