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

The correct option is A $14\Omega$ and $40\Omega$

Given, the current through resistance $20\Omega =1A$

Current through $R_2=0.5A$.

So, the voltage drop across the three parallel-connected resistance will be the same.

Therefore, across resistance $20\Omega$, voltage drop is

$V_P=IR,V_P=1\times 20=20V$.

That means, 20 V is dropped across the parallel combination.

so, Volage drop across $R_2=\frac{V_p}{0.5}=\frac{20}{0.5}$=$40\Omega$

Also, current across $10\Omega =\frac{V_p}{10}=\frac{20}{10}=2A$

so, remaining voltage drop will be across $R_1$,that is,

$V_1=69-V_P=69-20=49V$

Now, we know current is always same for components in series.

Therefore, current across the parallel combination will be the same as current across $R_1$.

so, current across $20\Omega =1A$

current across $10\Omega =2A$

current across $R_2=40\Omega =0.5A$

Total current across parallel combination,$I_P=1+2+0.5=3.5A$

So, $I_P=3.5A$ is current across $R_1$ also.

So, $R_1=\frac{V_1}{I_P}=\frac{49}{3.5}=14\Omega$

$R_1=14\Omega$
and $R_2=40\Omega$