Question

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

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

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 $$R_2= 40 \Omega$$                           Physics

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