Question

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

Answer 1

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,

\[2P' =  \left( \frac{220}{625 . 034} \right)^2  \times 0 . 034 \times 2\]

\[             = 0 . 0084  W = 8 . 4  mW\]