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प्रश्न
A bulb with rating 250 V, 100 W is connected to a power supply of 220 V situated 10 m away using a copper wire of area of cross-section 5 mm2. How much power will be consumed by the connecting wires? Resistivity of copper = 1.7 × 10−8 Ωm
उत्तर
Let R be the resistance of the bulb. If P is the power consumed by the bulb when operated at voltage V, then
\[R = \frac{V^2}{P} = \frac{\left( 250 \right)^2}{100} = 625 \Omega\]
Resistance of the copper wire,
\[R_c = \rho\frac{l}{A} = \frac{1 . 7 \times {10}^{- 8} \times 10}{5 \times {10}^{- 6}} = 0 . 034 \Omega\]
The effective resistance,
\[R_{eff} = R + R_c = 625 . 034 \Omega\]
The current supplied by the power station,
\[i = \frac{V}{R_{eff}} = \left\{ \frac{220}{625 . 034} \right\} A\]
The power supplied to one side of the connecting wire,
\[P' = i^2 R_c \]
\[ = \left( \frac{220}{625 . 034} \right)^2 \times 0 . 034\]
The total power supplied on both sides,
\[2P' = \left( \frac{220}{625 . 034} \right)^2 \times 0 . 034 \times 2\]
\[ = 0 . 0084 W = 8 . 4 mW\]
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