Question

Question 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.

Class 12 - Physics - Current Electricity Current Electricity Page-127

Answer 1

(a) if resistors are connected in parallel then , equivalent resistance can be found by
$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}+......$

here,R₁ = 2Ω , R₂ = 4Ω and R₃ = 5Ω
so, 1/Req = 1/2Ω + 1/4Ω + 1/5Ω
1/Req = (10 + 5 + 4)/20 = 19/20
so, Req = 20/19 = 1.05Ω

(b) potential of 20V will be same across each resistor , so currrent
$I_1=\frac{V}{R_1}=\frac{20}{2}=10A$

$I_2=\frac{V}{R_2}=\frac{20}{4}=5A$

$I_3=\frac{V}{R_3}=\frac{20}{5}=4A$

hence, total current drawn from the cell , $I=I_1+I_2+I_3=10A+5A+4A=19A$

Answer 2

answer

For parallel connection note / sign is divded by sign

• I/Rp= 1/R1 + 1/R2 + 1/ R3
• 1/ RP = 1/2 + 1/4 + 1/5
• 1/RP = 10 + 5 + 4 hole divided by 20
• RP = 20/9 ohms
• hope this help you