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Question

In the circuit shown in the given figure, the resistances $$R_{1}$$ and $$R_{2}$$ are respectively 

813204_73b41983316c4c5fa91b01ca6c024b25.png


A
14Ω and 40Ω
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B
40Ω and 14Ω
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C
40Ω and 30Ω
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D
14Ω and 30Ω
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Solution

The correct option is A $$14\Omega$$ and $$40\Omega$$
Given, the current through resistance $$20\Omega =1A$$
 Current through $$R_2=0.5 A$$.
So, the voltage drop across the three parallel-connected resistance will be the same.
Therefore, across resistance $$20\Omega$$, voltage drop is
               $$V_P=IR, V_P=1\times20=20V$$.
That means, 20 V is dropped across the parallel combination.
so, Volage drop across $$R_2= \dfrac{V_p}{0.5}=\dfrac{20}{0.5}$$=$$40\Omega$$

Also, current across $$10\Omega =\dfrac{V_p}{10}=\dfrac{20}{10}=2A$$
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, $$I_P=3.5 A $$ is current across $$R_1$$ also.
So, $$R_1=\dfrac{V_1}{I_P}=\dfrac{49}{3.5}=14\Omega$$
$$R_1=14\Omega$$
 and $$R_2= 40 \Omega$$

                           

1495394_813204_ans_9e22869ccb53443e82f36b1560b6f24a.png

Physics

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