Question

3 Resistors one of them having a resistance of 12Ω when connected in parallel together across a battery of 44v draws a current of 22A these resistors when connected in series across the same battery draws a current of 2A. Find the resistance value of the other 2 resistors?

• A. 8, 3 ohm’s
• B. 6, 4 ohm’s
• C. 12, 2 ohm’s
• D. 5, 5 ohm’s

Answer 1

The correct option is B

6, 4 ohm’s

Now in case of parallel set up the net resistance will be

$\frac{1}{12}$ + $\frac{1}{x}$ + $\frac{1}{y}$ = $\frac{12x+12y+xy}{12xy}$

$\frac{12xy}{12x+12y+xy}$ -----------------------(i)

Also by ohm’s law

V = IR

44 = 22R so R = 2Ω - - - - - - - - - - - - - - - - - (II)

(I) = (II) ⇒ 12xy = 24 (x + y) + 2xy

= 10xy = 24 (x + y) – - - - - - - - - - - - - - - - -(III)

In series setup

R_net = 12 + x + y - - - - - - - - - - - - - - - - - - - (IV)

By ohm’s law V = IR

44 = 2R ⇒ R = 22Ω - - - - - - - - - - - - - - - - - (V)

(IV) = (V) ⇒ 12 + x + y = 2c ⇒ x + y = 10 - - - - - - - -(VI)

Putting (VI) in (III) we get xy = 24 - - - - - - - - - - - - (VII)

Only option (b) satisfies both equation (VI) & (VII)