Power In Electric Circuits

To solve this problem, we need to analyze both circuits.

Circuit 1:

In Circuit 1, the voltmeter is connected in parallel with the resistor.

• The total resistance in the circuit is .
• The voltage across the resistor is equal to the source voltage, , as it is connected directly across the supply and resistor in series.
• Using Ohm's Law, the current through the circuit is calculated as:

Circuit 2:

In Circuit 2, the voltmeter is connected across another 10 Ω resistor in parallel with the supply.

• The total resistance across the supply is still effectively , considering parallel configuration does not change the effective resistance of any circuit in series with the supply.
• The voltage across the resistor is also equal to the source voltage, .
• The current can be determined similarly as:

Conclusion:

• The voltages and are equal: .
• The currents and are also equal: .

Thus, the correct answer is: and .

To find the ratio of power consumption by the bulbs in two different scenarios, we need to analyze the circuit configurations given in the problem.

1. Initial Setup (All Bulbs Glowing):
   • The circuit consists of 6 similar bulbs arranged in two parallel sections (A and B) with three bulbs each.
   • Each parallel section has 3 bulbs in series.
   • If the resistance of each bulb is R , the total resistance in each section is 3R .
   • Since the sections are parallel, the equivalent resistance R_1 of the entire circuit is: R_1 = \frac{3R \times 3R}{3R + 3R} = \frac{3R}{2} .
2. Power Consumption:
   • The power consumed is given by P = \frac{E^2}{R_1} , where E is the emf of the source.
   • Substituting R_1 : P_1 = \frac{E^2}{\frac{3R}{2}} = \frac{2E^2}{3R} .
3. Modified Setup (Two from Section A and One from Section B Glowing):
   • The circuit now consists of 2 bulbs in section A and 1 bulb in section B.
   • For Section A, the resistance becomes 2R and for Section B, it is R .
   • The equivalent resistance R_2 of the modified circuit is: R_2 = \frac{2R \times R}{2R + R} = \frac{2R}{3} .
4. Power Consumption:
   • The power consumed in this configuration is: P_2 = \frac{E^2}{R_2} = \frac{E^2}{\frac{2R}{3}} = \frac{3E^2}{2R} .
5. Ratio of Power Consumption:
   • We need to find the ratio \frac{P_1}{P_2} : \frac{P_1}{P_2} = \frac{\frac{2E^2}{3R}}{\frac{3E^2}{2R}} = \frac{2}{3} \times \frac{2R}{3E^2} \times \frac{3E^2}{2R} = \frac{9}{4} .

Therefore, the ratio of power consumption by the bulbs when all are glowing to the modified setup is 9:4.