Resistors $R_{1}$ and $R_{2}$ have equivalent resistant of 6 ohms when connected in the circuit shown below. The resistance of $R_{1}$ could be ( in ohms) :-

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 8
$\frac{R_{1} R_{2}}{R_{1} + R_{2}} = 6 Ω$

Apply hit and trial method:
Take $R_{1} = 1 Ω$ ;

$\frac{R_{2}}{1 + R_{2}} = 6 Ω$
$R_{2} = - 5 Ω$ (Not possible)

Take $R_{1} = 5 Ω$ ;

$\frac{5 R_{2}}{5 + R_{2}} = 6 Ω$
$R_{2} = - 30 Ω$ (Not possible)

Take $R_{1} = 8 Ω$ ;

$\frac{8 R_{2}}{8 + R_{2}} = 6 Ω$
$R_{2} = 6 Ω$ (Possible value of resistance)

Take $R_{1} = 4 Ω$ ;

$\frac{4 R_{2}}{4 + R_{2}} = 6 Ω$
$R_{2} = - 4 Ω$ (Not possible)

So, $R_{1} = 8 Ω$ is a possible value of resistance.

Suggest Corrections

Similar questions

R_{1}

R_{2}

6 Ω

R_{1}

Ω

(R_{1} < R_{2})

R_{1} = (100 + 3 -)

R_{2} = (200 + 4 -)

R_{1}, R_{2}

R_{3}

Join BYJU'S Learning Program