Answer to Question #196604 in Electric Circuits for Hamza

Question

Q.12.: Arrange three resistors each of 202 to obtain the equivalent resistances of (1) 1.52 (ii) 302 (111) 692

Also prove the result mathematically.

Answer 1

(i)

The minimum resistance with given three resistances of 202 Ohms each is the parallel combination of all three as given below:

The equivalent resistance R is given by:

R = R1 || R2 || R3

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3}$

$\frac{1}{R} = \frac{1}{202} + \frac{1}{202} + \frac{1}{202} = \frac{3}{202}$

Or R = 202/3 = 67.33 Ohms

This is the minimum possible equivalent resistance with given three resistance of 202 ohms each.

(ii)

In the given figure, the resistance R1 is in series with parallel combination of R2 and R3.

Therefore, the equivalent resistance R is given by:

R = R1 + (R2 || R3)

$Therefore, R = \frac{R2 * R3}{R2 + R3} = 202 + \frac{202 * 202}{202 + 202}$

On solving above, we get equivalence resistance as

R = 303 Ohms.

(iii)

When the three resistances are added in series, we get the maximum equivalent resistance. See below figure.

R1, R2 and R3 are in series, therefore, the equivalent resistance is given by:

R = R1 + R2 + R3 = 202 + 202 + 202 = 606 Ohms (This is the maximum equivalent resistance)

There is no possibility of getting total equivalent res. of 697 Ohms with three given resistance of 202 Ohms.