JEE PYQ: Current Electricity Question 53
Question 53 - 2019 (12 Apr 2019 Shift 1)
The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is:
(1) $6,\Omega$
(2) $8,\Omega$
(3) $1,\Omega$
(4) $16,\Omega$
Show Answer
Answer: (2)
Solution
Equivalent resistance: $R_{\text{eq}} = \frac{4R \times 4R}{4R + 4R} + R + \frac{6R \times 12R}{6R + 12R} + R = 2R + R + 4R + R = 8R$. $P = \frac{V^2}{R_{\text{eq}}}$, $4 = \frac{16^2}{8R}$, $R = \frac{256}{32} = 8,\Omega$.
BETA
