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QPages 218-219: Q5: HC Verma: Concepts of Physics - Volume 2 Thermal and Chemical Effects of Electric Current Pages 218-219: Q5 Solution: A bulb with rating 250 V, 100 W is connected to a power supply…

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 mm25 \text{ mm}^2. How much power will be consumed by the connecting wires? Resistivity of copper = 1.7×10−8 Ωm1.7 \times 10^{-8} \ \Omega\text{m}.

Step-by-Step Verified Solution

**Answer:** Power consumed by wires =8.4= 8.4 mW. **Step-by-Step Explanation:** Bulb resistance: Rb=2502/100=625 ΩR_b = 250^2/100 = 625\ \Omega. Wire resistance (one way): Rw=ρl/A=(1.7×10−8×10)/(5×10−6)=0.034 ΩR_w = \rho l/A = (1.7 \times 10^{-8} \times 10)/(5 \times 10^{-6}) = 0.034\ \Omega. Total wire resistance (two ways) =0.068 Ω= 0.068\ \Omega. Total resistance =625.068 Ω= 625.068\ \Omega. Current =220/625.068=0.352= 220/625.068 = 0.352 A. Power in wires =i2Rw=(0.352)2×0.068=0.0084= i^2 R_w = (0.352)^2 \times 0.068 = 0.0084 W .

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