JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 25)
A uniform heating wire of resistance 36$$\Omega$$ is connected across a potential difference of 240 V. The wire is then cut into half and potential difference of 240V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1 : x, where x is ____________
Answer
4
Explanation
For Case I,
The potential difference of the uniform wire, V = 240 V
The resistance of the uniform wire, R1 = 36 $$\Omega$$
The power dissipation in the first case,
$${P_1} = {{{V^2}} \over {{R_1}}} = {{{{(240)}^2}} \over {36}}$$
For Case II,
The resistance of each half, $${R_2} = {{{R_1}} \over 2} = {{36} \over 2} = 18\Omega $$
$${P_2} = {{{V^2}} \over {{R_2}}} + {{{V^2}} \over {{R_2}}} = {{{{(240)}^2}} \over {18}} + {{{{(240)}^2}} \over {18}} = {{{{(240)}^2}} \over 9}$$
Thus, the ratio of the total power dissipation in the first case to the second case
$${{{P_1}} \over {{P_2}}} = {{{{(240)}^2}/36} \over {{{(240)}^2}/9}} \Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over 4}$$
Comparing with, $${{{P_1}} \over {{P_2}}} = {1 \over x}$$
The value of the x = 4.
The potential difference of the uniform wire, V = 240 V
The resistance of the uniform wire, R1 = 36 $$\Omega$$
The power dissipation in the first case,
$${P_1} = {{{V^2}} \over {{R_1}}} = {{{{(240)}^2}} \over {36}}$$
For Case II,
The resistance of each half, $${R_2} = {{{R_1}} \over 2} = {{36} \over 2} = 18\Omega $$
$${P_2} = {{{V^2}} \over {{R_2}}} + {{{V^2}} \over {{R_2}}} = {{{{(240)}^2}} \over {18}} + {{{{(240)}^2}} \over {18}} = {{{{(240)}^2}} \over 9}$$
Thus, the ratio of the total power dissipation in the first case to the second case
$${{{P_1}} \over {{P_2}}} = {{{{(240)}^2}/36} \over {{{(240)}^2}/9}} \Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over 4}$$
Comparing with, $${{{P_1}} \over {{P_2}}} = {1 \over x}$$
The value of the x = 4.
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