Ncert Exercises & Solutions

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.

(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,

$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$
$= \frac{1}{2} + \frac{1}{4} + \frac{1}{5} = \frac{10 + 5 + 4}{20} = \frac{19}{20}$
$∴ R = \frac{20}{19} Ω$

Therefore, total resistance of the combination is $\frac{20}{19} Ω$ .

(b) Emf of the battery, V = 20 V

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

$I_{1} = \frac{V}{R_{1}} = \frac{20}{2} = 10 A$

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

$I_{2} = \frac{V}{R_{2}} = \frac{20}{4} = 5 A$

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

$I_{3} = \frac{V}{R_{3}} = \frac{20}{5} = 4 A$

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

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

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