Question 211:
easyTwo $220\text{ volt}$, $100\text{ watt}$ bulbs are connected first in series and then in parallel. Each time the combination is connected to a $220\text{ volt}$ a.c. supply line. The power drawn by the combination in each case respectively will be:
(2003)
In series, total power is $P_{\text{series}} = \frac{P_1 P_2}{P_1 + P_2} = \frac{100 \times 100}{200} = 50\text{ W}$. In parallel, total power is $P_{\text{parallel}} = P_1 + P_2 = 100 + 100 = 200\text{ W}$.
Question 212:
easyTwo bulbs of ($40\text{ W}$, $200\text{ V}$), and ($100\text{ W}$, $200\text{ V}$). Then correct relation for their resistances:
(2000)
Since resistance $R = \frac{V^2}{P}$ and both have the same voltage rating, $R$ is inversely proportional to $P$. The $40\text{ W}$ bulb has lower power, so it must have higher resistance: $R_{40} > R_{100}$.
Question 213:
easyWhen three identical bulbs are connected in series, the consumed power is $10\text{ W}$. If they are now connected in parallel then the consumed power will be:
(1998)
In series, $P_{\text{series}} = \frac{P}{3} = 10\text{ W}$, so the power of each bulb is $P = 30\text{ W}$. When connected in parallel, $P_{\text{parallel}} = 3P = 3 \times 30 = 90\text{ W}$.
Question 214:
easyA ($100\text{ W}$, $200\text{ V}$) bulb is connected to a $160\text{ volts}$ supply. The power consumption would be
(1997)
Resistance of the bulb $R = \frac{V^2}{P} = \frac{200^2}{100} = 400\Omega$. The power consumed at $160\text{ V}$ is $P' = \frac{V'^2}{R} = \frac{160^2}{400} = 64\text{ W}$.
Question 215:
easyIf two bulbs, whose resistances are in the ratio of $1 : 2$ are connected in series, the power dissipated in them has the ratio of
(1997)
In a series circuit, the current $I$ is the same through both bulbs. Since power $P = I^2 R$, the power dissipated is directly proportional to the resistance. Hence, the ratio is $1 : 2$.
Question 216:
easyWhen three identical bulbs of $60\text{ watt}$, $200\text{ volt}$ rating are connected in series to a $200\text{ volt}$ supply, the power drawn by them will be: (2004)
Resistance of one bulb is $R = \frac{V^2}{P} = \frac{200^2}{60}$. In series, total resistance $R_{\text{eq}} = 3R$. Power drawn is $P' = \frac{V^2}{3R} = \frac{1}{3} \frac{V^2}{R} = \frac{60}{3} = 20\text{ watt}$.
Question 217:
easyTwo cities are $150\text{ km}$ apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is $8\text{ volt}$ and the average resistance per km is $0.5\Omega$. The power loss in the wire is:
(2014)
Total resistance $R = 150 \times 0.5 = 75\Omega$. Total voltage drop $V = 150 \times 8 = 1200\text{ V}$. Power loss $P = \frac{V^2}{R} = \frac{(1200)^2}{75} = 19200\text{ W} = 19.2\text{ kW}$.
Question 218:
easyIf voltage across a bulb rated $220\text{ Volt} – 100\text{ Watt}$ drops by $2.5\%$ of its rated value, the percentage of the rated value by which the power would decrease is:
(2012 Pre)
Power is given by $P = \frac{V^2}{R}$. For small changes, the fractional change is $\frac{\Delta P}{P} = 2\frac{\Delta V}{V}$. The percentage decrease in power will be $2 \times 2.5\% = 5\%$.
Question 219:
easyIn India electricity is supplied for domestic use at $220\text{ V}$. It is supplied at $110\text{ V}$ in USA. If the resistance of a $60\text{ W}$ bulb for use in India is $R$, the resistance of a $60\text{ W}$ bulb for use in USA will be:
2004
Power $P = \frac{V^2}{R}$, which implies $R \propto V^2$ for the same power. Thus, $\frac{R_{\text{USA}}}{R_{\text{India}}} = (\frac{110}{220})^2 = \frac{1}{4}$, yielding $R_{\text{USA}} = \frac{R}{4}$.
Question 220:
easyTwo $220\text{ volt}$, $100\text{ watt}$ bulbs are connected first in series and then in parallel. Each time the combination is connected to a $220\text{ volt}$ a.c. supply line. The power drawn by the combination in each case respectively will be:
(2003)
In series, total power is $P_{\text{series}} = \frac{P_1 P_2}{P_1 + P_2} = \frac{100 \times 100}{200} = 50\text{ W}$. In parallel, total power is $P_{\text{parallel}} = P_1 + P_2 = 100 + 100 = 200\text{ W}$.