Question

In the circuit shown, the galvanometer shows zero current.The value of resistance R is :

• A. $1\Omega$
• B. $10\Omega$
• C. $4\Omega$
• D. $9\Omega$

Answer 1

Answer: (B)

$10\Omega$

since no current is flowing through galvanometer, the potential drop across R is 8V.

Hence we can write

$\left(\frac{12}{R+5}\right)R=8$

$12R=8R+40$

$\therefore R=10\Omega$