Chapter 11

NCERT
Class 10
Science
Solutions
4. How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω

Question:

How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω

Answer:

(a) To get 4 Ω equivalent resistance, 3 Ω and 6 Ω resistors are connected in parallel and 2 Ω in series.

When 3 Ω and 6 Ω are added in parallel.

1/RP = 1/R1 + 1/R2

1/RP = 1/3 + 1/6

1/RP = (2+1)/6

1/RP = 3/6

        = 2Ω

When its equivalent resistance is connected in series with a resistor of 2 Ω, then

RS = R1 + R2

RS = 2 + 2

    = 4Ω

When all three resistors are connected, the equivalent resistance is 1 Ω.

1/RP = 1/R1 + 1/R2 + 1/R3

1/RP = (3 +2 + 1)/6

1/RP = 6/6

      = 1/1

RP   = 1Ω

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