Study the circuit shown in which two resistors X and Y of resistances $3 \Omega$ and $6 \Omega$ respectively are joined in series with a battery of $2 \text{ V}$. (I) Draw a circuit diagram showing the above two resistors X and Y joined in parallel with same battery and same ammeter and voltmeter. (II) In which combination of resistors will the (i) potential difference across X and Y and (ii) current through X and Y, be the same? (III) (a) Find the current drawn from the battery by the series combination of the two resistors (X and Y). OR (III) (b) Determine the equivalent resistance of the parallel combination of the two resistors (X and Y).

Quick Oral Answer

The circuit diagram for the parallel combination of resistors X and Y with the same battery and ammeter and voltmeter shows that both resistors have the same potential difference across them, which is equal to the voltage of the battery. The current through each resistor is different, depending on its resistance. The equivalent resistance of the parallel combination can be calculated using the formula: 1/R_eq = 1/R_X + 1/R_Y. In this case, R_X = 3 ohms and R_Y = 6 ohms, so the equivalent resistance is: 1/R_eq = 1/3 + 1/6 = 1/2 ohms. Therefore, the equivalent resistance of the parallel combination is 2 ohms. To find the current drawn from the battery by the series combination of the two resistors, we can use Ohm's law: I = V/R, where I is the current, V is the voltage, and R is the total resistance. The total resistance of the series combination is the sum of the individual resistances: R_total = R_X + R_Y = 3 + 6 = 9 ohms. The voltage of the battery is 2 volts, so the current drawn from the battery is: I = 2/9 = 0.22 amps. Therefore, the current drawn from the battery by the series combination of the two resistors is 0.22 amps.