(a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistance joined in parallel is equal to the sum of the reciprocal of the individual resistances. (b) In an electric circuit two resistors of 12Ω each are joined in parallel to a 6V battery. Find the voltage drawn from the battery.

(a) Proof for parallel resistors:

Consider three resistors R₁, R₂, and R₃ connected in parallel across a voltage source V. Let the equivalent resistance of this parallel combination be Rₑ.

[Insert a circuit diagram here showing three resistors R₁, R₂, R₃ connected in parallel across a voltage source V. The equivalent resistance Rₑ should be shown in parallel with the voltage source.]

According to Ohm's law, the current flowing through each resistor is:

I₁ = V/R₁
I₂ = V/R₂
I₃ = V/R₃

The total current I supplied by the source is the sum of the individual currents:

I = I₁ + I₂ + I₃
I = V/R₁ + V/R₂ + V/R₃
I = V(1/R₁ + 1/R₂ + 1/R₃)

By Ohm's law, the total current is also given by:

I = V/Rₑ

Therefore, we can equate the two expressions for I:

V/Rₑ = V(1/R₁ + 1/R₂ + 1/R₃)

Dividing both sides by V (assuming V ≠ 0), we get:

1/Rₑ = 1/R₁ + 1/R₂ + 1/R₃

This proves that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.

(b) Calculation for parallel resistors:

Two resistors of 12Ω each are connected in parallel across a 6V battery.

The equivalent resistance Rₑ is given by:

1/Rₑ = 1/12Ω + 1/12Ω = 2/12Ω = 1/6Ω

Therefore, Rₑ = 6Ω

The voltage drawn from the battery is the voltage across the parallel combination, which is 6V. Since the resistors are in parallel, the voltage across each resistor is equal to the voltage of the battery.

Therefore, the voltage drawn from the battery is 6V.