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?

Asked by Vishal kumar | 1 year ago |  205

##### Solution :-

(a) The circuit diagram below shows the connection of three resistors

From the circuit above, it is understood that 3 Ω and 6 Ω are connected in parallel. Hence, their equivalent resistance is given by

$$\frac{1}{R}=\frac{1}{3}+\frac{1}{6}$$

$$R= \frac{1}{\frac{1}{6}+\frac{1}{3}}=\frac{6\times3}{6+3}=2\,\Omega$$

The equivalent resistor 2 Ω is in series with the 2 Ω resistor. Now the equivalent resistance can be calculated as follows:

Req= 2 Ω +2 Ω = 4 Ω

Hence, the total resistance of the circuit is 4 Ω.

(b) The circuit diagram below, shows the connection of three resistors

From the circuit, it is understood that all the resistors are connected in parallel. Therefore, their equivalent resistance can be calculated as follows:

$$R=\frac{1}{\frac{1+1+1}{2+3+6}} \frac{1}{\frac{3+2+1}{6}}=\frac{6}{6}=1\,\Omega$$

The total resistance of the circuit is 1 Ω.

Answered by Shivani Kumari | 1 year ago

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