Two conducting rods A and B of same length and cross-sectional area are connected (i) In series (ii) In parallel as shown. In both combination a temperature difference of 100°C is maintained. If thermal conductivity of A is 3K and that of B is K then the ratio of heat current flowing in parallel combination to that flowing in series combination is
(a) $\frac{16}{3}$ (b) $\frac{3}{16}$ (c) $\frac{1}{1}$ (d) $\frac{1}{3}$
Text Solution
Verified by ExpertsA
$Heat current H = \frac{\Delta \theta}{R} \Rightarrow \frac{H_P}{H_S} = \frac{R_S}{R_P} In first case: R_S = R_1 + R_2 = \frac{1}{(3K)A} + \frac{1}{KA} = \frac{4}{3} \frac{1}{KA} In second case: R_P = \frac{R_1 R_2}{R_1 + R_2} = \frac{\frac{1}{(3K)A} \times \frac{1}{KA}}{\left(\frac{1}{(3K)A} + \frac{1}{KA}\right)} = \frac{1}{4KA} \therefore \frac{H_P}{H_S} = \frac{\frac{4l}{3KA}}{\frac{l}{4KA}} = \frac{16}{3}$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems