QJEEM21Ph61a4: JEE Main 2021 Physics - Class 12 JEEM21Ph61a4 Solution: The voltage across the 10 resistor in the given circuit is x…

The voltage across the 10 $\Omega$ resistor in the given circuit is x volt.


The value of 'x' to the nearest integer is _________.

Step-by-Step Verified Solution

${R_{eq}} = {{50 \times 20} \over {50 + 20}}$

${R_{eq}} = {{1000} \over {70}} = {{100} \over 7}\Omega$



${R_{ef}} = 10 + {{100} \over 7} = {{170} \over 7}\Omega$



$I = {V \over R} = {{170/7} \over {170}} = {{170} \over {170/7}}$

I = 7A

Potential Across 10 $\Omega$ resister

V₁₀ = IR = 7 $\times$ 10 = 70V

$\Rightarrow$ V₁₀ = 70V