(a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

Ans (a).
There are three resistors of resistances,

R₁ = 2 Ω, R₂ = 4 Ω, and R₃ = 5 Ω

They are connected in parallel. Hence, total resistance (R) of the combination is given by,

1/R =1/R₁ + 1/R₂ +1/R₃

= 1/2 + 1/4 + 1/5 = (10+5+4 )/20 = 19/20

Therefore R = 20/19 Ω

Therefore, total resistance of the combination is 20/19 Ω.

Ans (b).

Emf of the battery, V = 20 V

Current (I₁) flowing through resistor R₁ is given by,

I₁ =V/R₁ = 20/2 =10 A

Current (I₂) flowing through resistor R₂ is given by,

I₂ =V/R₂= 20/4 = 5A

Current (I₃) flowing through resistor R₃ is given by,

I₃=V/ R₃ =20/5 =4 A

Total current, I = I₁ + I2 + I3 = 10 + 5 + 4 = 19 A

Therefore, the current through each resister is 10 A, 5 A, and 4 A respectively and the total current is 19 A.