JEE PYQ: Current Electricity Question 11
Question 11 - 2021 (18 Mar 2021 Shift 1)
The voltage across the $10,\Omega$ resistor in the given circuit is x volt.
[Circuit: $10,\Omega$ in parallel with series combination of $50,\Omega$ and $20,\Omega$, connected to 170 V source]
The value of ‘x’ to the nearest integer is ____.
Show Answer
Answer: 70
Solution
$R_{\text{eq}} = \frac{50 \times 20}{70} = \frac{100}{7}$. Parallel with 10: $R_{\text{eq}} = \frac{170}{7} \div (\frac{170}{7} + 170) \times 170$. $v_1 = \frac{170/7}{170/7} \times 10 = 70$ V.
BETA
