JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 16)
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:
_12th_April_Morning_Slot_en_16_1.png)
16 $$\Omega $$
1 $$\Omega $$
8 $$\Omega $$
6 $$\Omega $$
Explanation
Req = 2R + R + 4R + R = 8R
$$P = {{{v^2}} \over {{R_{eq}}}} \Rightarrow {{16 \times 16} \over {8R}} = 4\,watt$$
$${{16 \times 16} \over {4 \times 8}} = R \Rightarrow R = 8\Omega $$
$$P = {{{v^2}} \over {{R_{eq}}}} \Rightarrow {{16 \times 16} \over {8R}} = 4\,watt$$
$${{16 \times 16} \over {4 \times 8}} = R \Rightarrow R = 8\Omega $$
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