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Electrical power
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Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volt and the average resistance per km is 0.5 $\Omega$. The power loss in the wire is:Resistance = (0.5 $\Omega$/km)(150 km) = 75 $\Omega$
Total voltage drop = (8 V/km)(150 km) = 1200 V
Power loss $= \frac{(\Delta V)^2}{R} = \frac{(1200)^2}{75} = 19200$ W = 19.2 kW -
If voltage across a bulb rated 220 Volt-100 Watt drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is$P = \frac{V^2}{R}$, with R constant, so $P \propto V^2$
$\frac{\Delta P}{P}\times100 = 2\times\frac{\Delta V}{V}\times100$
$= 2\times2.5 = 5\%$ -
The power dissipated in the circuit shown in the figure is 30 Watts. The value of R is
R and the 5 $\Omega$ resistor are in parallel across the 10 V battery.
Total power: $\frac{V^2}{5} + \frac{V^2}{R} = 30 \Rightarrow \frac{(10)^2}{5} + \frac{(10)^2}{R} = 30$
$20 + \frac{100}{R} = 30 \Rightarrow \frac{100}{R} = 10$
$R = 10\ \Omega$