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 Options: A . &nbsp 16 3 B . &nbsp 3 16 C . &nbsp 1 1 D . &nbsp 1 3

Answer: Option A : A Heat current H = Δ θ R ⇒ H p H s = R s R p In first case : R s = R 1 + R 2 = l ( 3 K ) A + l K A = 4 3 l K A In second case : R p = R 1 R 2 R 1 + R 2 = 1 ( 3 K ) A × l K A ( l ( 3 K ) A + l K A ) = l 4 K A ∴ H p H s = 4 l 3 K A l 4 K A = 16 3

Two conducting rods A and B of same length and cross-sectional area are connecte

Discussion Forum : Heat Transfer

Question -

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 $100^{\circ}$ 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

Options:

A .

\frac{16}{3}

B .

\frac{3}{16}

C .

\frac{1}{1}

D .

\frac{1}{3}

Answer: Option A
:
A

Heat current $H = \frac{Δ θ}{R} \Rightarrow \frac{H_{p}}{H_{s}} = \frac{R_{s}}{R_{p}}$
In first case : $R_{s} = R_{1} + R_{2} = \frac{l}{(3 K) A} + \frac{l}{K A} = \frac{4}{3} \frac{l}{K A}$
In second case : $R_{p} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{\frac{1}{(3 K) A} \times \frac{l}{K A}}{(\frac{l}{(3 K) A} + \frac{l}{K A})} = \frac{l}{4 K A}$
$∴ \frac{H_{p}}{H_{s}} = \frac{\frac{4 l}{3 K A}}{\frac{l}{4 K A}} = \frac{16}{3}$

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