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VITEEE 2010 Paper

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Question : 2 of 120

Marks: +1, -0

The ratio of the amounts of heat developed in the four arms of a balance Wheatstone bridge, when the arms have resistances P = 100Ω, Q = 10Ω, R = 300Ω and S = 30Ω respectively is

3:30:1:10
30:3:10:1
30:10:1:3
30:1:3:10

Solution:

Current through arms of resistances

P

and

Q

in series

i_{1}=\frac{i \times 330}{330+110}

=

\frac{3}{4}i

Here

i

= total current

Similarly, current through arms of resistances

R

and

S

in series

i_{2}=\frac{i \times 110}{330+110}

=

\frac{1}{4}i

Heat developed per second =

i^{2}R

∴ Ratio of heat developed per sec

H_{P}:H_{Q}:H_{R}:H_{S}

=

\left(\frac{3}{4}i\right)^{2} \times 100:\left(\frac{3}{4}i\right)^{2} \times 10:\left(\frac{1}{4}i\right)^{2} \times 300

:\left(\frac{1}{4}i\right)^{2} \times 30

= 30:3:10:1

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