Class 12 Physics
CBSE and UP Board
Current Electricity
NCERT Solutions for Class 12 Physics Chapter 3
Chapter-3 Exercise 3.4
(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.
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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.