Question

In the circuit shown in the given figure, the resistances $R_1$ and $R_2$ are respectively

• A. $14 \,\Omega$ and $40 \,\Omega$
• B. $40 \,\Omega$ and $14 \,\Omega$
• C. $40 \,\Omega$ and $30 \,\Omega$
• D. $14 \,\Omega$ and $30 \,\Omega$

Answer 1

Answer: (A)

Potential difference across $20 \,\Omega = 20 \times 1$
$= 20 \, V$ = potential difference across $R_2$
Current in $R_2 = 0.5 \, A$
$\therefore \:\: R_{2 } = \frac{20 V}{0.5 A} = 40 \Omega$
Potential difference across $R_1 = 69 \, V - 20 \, V $
$=49 \, V$
Current in $R_1 = 0.5 A + \frac{20}{10} A + \, A = 3.5 \, A}$
$\therefore \:\:\: R_1 = \frac{49}{3.5} = 14 \, \Omega$