Class 12 Physics
CBSE and UP Board
Current Electricity
Additional Exercise
Chapter-3 Exercise 3.32
NCERT Solutions Class 12 Physics Chapter 3 Question 32
Determine the current drawn from a 12V supply with internal resistance 0.5Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.
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The resistance of each resistor connected in the given circuit,
R = 1 Ω
Equivalent resistance of the given circuit = R’
The network is infinite.
Hence, equivalent resistance is given by the relation,
R’ = 2 + R’/(R+1)
(R’)² -2 R’ -2 = 0
R’ = [2 ± √ (4+8)]/2
= [2 ± √12]/2 = 1 ± √3
Negative value of R’ cannot be accepted.
Hence, equivalent resistance,
R’=(1 + √3)= 1 + 1.73 = 2.73 Ω
Internal resistance of the circuit, r = 0.5 Ω
Hence, total resistance of the given circuit
= 2.73 + 0.5 = 3.23 Ω
Supply voltage, V = 12 V
According to Ohm’s Law, current drawn from the source is given by the ratio, 12/3.23= 3.72 A