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JEE Main 2021 (18 Mar, Shift 1) · Kirchhoff's Laws & Measuring Instruments

Kirchhoff's Laws & Measuring Instruments question from JEE Main 2021 (18 Mar, Shift 1)

Q126JEE Main 2021 · 18 MAR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments

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

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

Numerical answer type. Work it out, then open the answer.

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Answer: 70

$50\ \Omega \| 20\ \Omega = \dfrac{50 \times 20}{70} = \dfrac{100}{7}\ \Omega$
Total $R = 10 + \dfrac{100}{7} = \dfrac{170}{7}\ \Omega$ , so $I = \dfrac{170}{170/7} = 7\text{ A}$ .
Voltage across $10\ \Omega$ : $x = 7 \times 10 = 70\text{ V}$
$x = 70$

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